linear algebra practice problems and solutions

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Linear algebra

Unit 1: vectors and spaces, unit 2: matrix transformations, unit 3: alternate coordinate systems (bases).

Linear Algebra Questions

Linear algebra questions with solutions are provided here for practice and to understand what is linear algebra and its application to solving problems. Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear functions which operate on vectors and follow vector addition.

Following are the main topics under linear algebra:

• Matrices and determinants
• Vector Spaces
• System of linear equations
• Linear transformations
• Inner product spaces
• Diagonalizations and quadratic forms

We shall practice a few problems based on these topics.

Learn more about linear algebra and its applications .

Linear Algebra Questions with Solutions

Let us solve a few questions based on linear algebra.

Question 1:

Show that the matrix A is unitary matrix

\(\begin{array}{l}A=\frac{1}{5}\begin{bmatrix}-1+2i & -4-2i \\ 2-4i& -2-i \\\end{bmatrix}\end{array} \)

A matrix is said to be unitary if and only if AA* = A*A = I, where A* is the transpose of the conjugate of A.

Transpose of A

\(\begin{array}{l}A^{T}=\frac{1}{5}\begin{bmatrix}-1+2i & 2-4i \\ -4-2i& -2-i \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}A^{*}=\overline{A^{T}}=\frac{1}{5}\begin{bmatrix}-1-2i & 2+4i \\ -4+2i& -2+i \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}AA^{*}=\frac{1}{25}\begin{bmatrix}-1+2i & -4-2i \\2-4i & -2-i \\\end{bmatrix}\begin{bmatrix}-1-2i & 2+4i \\ -4+2i& -2+i \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}=\frac{1}{25}\begin{bmatrix}1+4+16+4 & 0\\0 & 4+16+4+1 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}\therefore AA^{*}=\begin{bmatrix}1 & 0 \\0 & 1 \\\end{bmatrix}=I\end{array} \)

Similarly, we can show that A*A = I

Hence, A is a unitary matrix.

Also refer: Types of Matrices

Question 2:

Find the rank and the nullity of the following matrix:

\(\begin{array}{l}\begin{bmatrix}1 & -2 & -1 & 4 \\2 & -4 & 3 & 5 \\-1 & 2 & 6 & -7 \\\end{bmatrix}\end{array} \)

To find the rank and nullity of the given matrix, we transform the given matrix into a row-reduced echelon form, by performing elementary transformations.

\(\begin{array}{l}A=\begin{bmatrix}1 & -2 & -1 & 4 \\2 & -4 & 3 & 5 \\-1 & 2 & 6 & -7 \\\end{bmatrix}\end{array} \)

Applying R 2 → R 2 – 2R 1 and R 3 → R 3 + R 1

\(\begin{array}{l}A~\begin{bmatrix}1 & -2 & -1 & 4 \\0 & 0 & 5 & -3 \\0 & 0 & 5 & -3 \\\end{bmatrix}\end{array} \)

Applying C 2 → C 2 + 2C 1 , C 3 → C 3 + C 1 and C 4 → C 4 – 4C 1

\(\begin{array}{l}A~\begin{bmatrix}1 & 0 & 0 & 0 \\0 & 0 & 5 & -3 \\0 & 0 & 5 & -3 \\\end{bmatrix}\end{array} \)

Applying R 3 → R 3 – R 2

\(\begin{array}{l}A~\begin{bmatrix}1 & 0 & 0 & 0 \\0 & 0 & 5 & -3 \\0 & 0 & 0 & 0 \\\end{bmatrix}\end{array} \)

Applying R 2 → (⅕)R 2

\(\begin{array}{l}A~\begin{bmatrix}1 & 0 & 0 & 0 \\0 & 0 & 1 & -3/5 \\0 & 0 & 0 & 0 \\\end{bmatrix}\end{array} \)

Applying C 4 → C 4 + (⅗)C 3

\(\begin{array}{l}A~\begin{bmatrix}1 & 0 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 0 \\\end{bmatrix}\end{array} \)

∴ number of non-zero rows of the row-reduced echelon form of A = rank of A = 2

number of zero rows of the row-reduced echelon form of A = nullity of A = 2

Learn more about rank and nullity .

Question 3:

Solve the following system of linear equations:

x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30

The given linear equations can be written in the form of a matrix equation AX = B, where

\(\begin{array}{l}A=\begin{bmatrix}1 & 1 & 1 \\1 & 2 & 3 \\1 & 4 & 7 \\\end{bmatrix}, X = \begin{bmatrix}x \\y \\z\end{bmatrix}\:\:and\:\:B=\begin{bmatrix}6 \\14 \\30\end{bmatrix}\end{array} \)

The augmented matrix [A| B] is-

\(\begin{array}{l}[A|B]=\begin{bmatrix}1 & 1 & 1 &|6 \\1 & 2 & 3&|14 \\1 & 4 & 7 &|30 \\\end{bmatrix}\end{array} \)

We reduce the given matrix to row echelon form by applying elementary row transformations

Applying R 2 → R 2 – R 1 , R 3 → R 3 – R 1

\(\begin{array}{l}[A|B]\sim \begin{bmatrix}1 & 1 & 1 &|6 \\0 & 1& 2&|8 \\0 & 3 & 6&|24 \\\end{bmatrix}\end{array} \)

Applying R 3 → R 3 – 3R 2

\(\begin{array}{l}[A|B]\sim \begin{bmatrix}1 & 1 & 1 &|6 \\0 & 1& 2&|8 \\0 & 0 & 0&|0 \\\end{bmatrix}\end{array} \)

Since, Rank of A = Rank of [A : B] = 2 < number of unknowns

∴ the given system of linear equations has an infinite number of solutions.

Thus, we get from the row reduced echelon form matrix

x + y + z = 6 ….(i)

⇒ y = 8 – 2z putting this value of y in (i), we get

x + 8 – 2z + z = 6

⇒ x – z = –2

⇒ x = z – 2

Now taking different values of z will give different values of the given system of equations.

• Transpose of Matrix
• Determinant of a Matrix
• Matrix Multiplication
• Matrix Operations
• Special Matrices

Question 4:

Show that the set V = {(x, y) ∈ R 2 | xy ≥ 0} is not a vector space of R 2 .

For V to be a vector space, it is required that V must be closed under addition, that is for any x and y in V, x + y ∈ V

Let ( – 1, 0) and (0, 1) ∈ V

Now, ( – 1, 0) + (0, 1) = ( –1 + 0, 0 + 1) = ( –1, 1)

But, –1 × 1 = –1 < 0 ⇒ ( –1, 1) ∉ V.

∴ V is not a vector space in R 2 .

Question 5:

Find the eigenvalues of

\(\begin{array}{l}A= \begin{bmatrix}0 & 1 & 0 \\0 & 0 & 1 \\4 & -17 & 8 \\\end{bmatrix}\end{array} \)

The characteristic polynomial is given by

\(\begin{array}{l}det(A-\lambda I) = det\begin{bmatrix}-\lambda & -1&0 \\ 0& -\lambda & 1 \\4 & -17 & 8-\lambda \\\end{bmatrix}=\lambda ^{3}-8\lambda^{2}+17\lambda-4\end{array} \)

Eigenvalues of A are the roots of the above cubic equation,

𝜆 3 – 8𝜆 2 + 17𝜆 – 4 = 0

⇒ (𝜆 – 4)(𝜆 2 – 4𝜆 + 1) = 0

Solving this we get,

𝜆 = 4, 𝜆 = 2 ±√3

These are the eigenvalues of A.

Also check: Eigenvalues and Eigenvectors

Determine whether the following vector is linearly dependent or linearly independent: (1, 2, –3, 1), (3, 7, 1, –2), (1, 3, 7, –4).

The vectors could form the column vectors of matrix A. We shall find the rank of A by reducing it to row echelon form.

\(\begin{array}{l}A=\begin{bmatrix}1 & 3 & 1 \\2 & 7 & 3 \\-3 & 1 & 7 \\1 & -2 & -4 \\\end{bmatrix}\end{array} \)

Applying R 2 → R 2 – 2R 1 , R 3 → R 3 + 3R 1 , and R 4 → R 4 – R 1

\(\begin{array}{l}A\sim \begin{bmatrix}1 & 3 & 1 \\0 & 1 & 1 \\0 & 10 & 10 \\0 & -5 & -5 \\\end{bmatrix}\end{array} \)

Applying R 3 → R 3 – 10R 2 , R 4 → R 4 + 5R 2

\(\begin{array}{l}A\sim \begin{bmatrix}1 & 3 & 1 \\0 & 1 & 1 \\0 & 0 & 0 \\0 & 0 & 0 \\\end{bmatrix}\end{array} \)

Clearly, rank of A = 2 < number of column vectors. So, the given vectors are linearly dependent.

Question 6:

Verify whether the polynomials x 3 – 5x 2 – 2x + 3, x 3 – 1, x 3 + 2x + 4 are linearly independent.

We may construct a matrix with coefficients of x 3 , x 2 , x, and constant terms.

\(\begin{array}{l}A=\begin{bmatrix}1 & 1 & 1 \\-5 & 0 & 0 \\-2 & 0 & 2 \\3 & -1 & 4 \\\end{bmatrix}\end{array} \)

To find the rank of A let us reduce it to row echelon form by applying elementary transformations

Applying R 2 → R 2 + 5R 1 , R 3 → R 3 + 2R 1 , and R 4 → R 4 – 3R 1

\(\begin{array}{l}A\sim \begin{bmatrix}1 & 1 & 1 \\0 & 5 & 5 \\0 & 2 & 4 \\0 & -4 & 1 \\\end{bmatrix}\end{array} \)

Applying R 2 → (⅕) R 2

\(\begin{array}{l}A\sim \begin{bmatrix}1 & 1 & 1 \\0 & 1 & 1 \\0 & 2 & 4 \\0 & -4 & 1 \\\end{bmatrix}\end{array} \)

Applying R 3 → R 3 – 2R 2 , R 4 → R 4 + 4R 2

\(\begin{array}{l}A\sim \begin{bmatrix}1 & 1 & 1 \\0 & 1 & 1 \\0 & 0 & 2 \\0 & 0 & 5 \\\end{bmatrix}\end{array} \)

Applying R 4 → R 4 – (5/2)R 3

\(\begin{array}{l}A\sim \begin{bmatrix}1 & 1 & 1 \\0 & 1 & 1 \\0 & 0 & 2 \\0 & 0 & 0 \\\end{bmatrix}\end{array} \)

∴ rank of A = 3 = number of column vectors. So the given vectors are linearly independent.

Question 7:

Show that the following matrix is diagonalizable:

\(\begin{array}{l}A=\begin{bmatrix}1 & 0 & -1 \\1 & 2 & 1 \\2 & 2 & 3 \\\end{bmatrix}\end{array} \)

First, we shall find the eigenvalues of A. The characteristic equation of A is given by:

\(\begin{array}{l}|A-\lambda I|=\begin{vmatrix}1-\lambda & 0 & -1 \\1 & 2-\lambda & 1 \\2 & 2 & 3-\lambda \\\end{vmatrix}=0\end{array} \)

⇒ (1 – 𝜆)(2 – 𝜆)(3 – 𝜆) = 0

⇒ 𝜆 = 1, 2, 3.

The eigenvector corresponding to 𝜆 1 = 1 is the non-zero solution of the following matrix equation:

(A – 1I)X = 0

\(\begin{array}{l}\Rightarrow \begin{bmatrix}0 & 0 & -1 \\1 & 1 & 1 \\2 & 2 & 2 \\\end{bmatrix}\begin{bmatrix}x \\y \\z\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

Applying elementary transformation R 3 → R 3 – 2R 2 and R 2 → R 2 + R 1 , we get

\(\begin{array}{l}\Rightarrow \begin{bmatrix}0 & 0 & -1 \\1 & 1 & 0 \\0 & 0 & 0 \\\end{bmatrix}\begin{bmatrix}x \\y \\z\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

\(\begin{array}{l}\Rightarrow \begin{bmatrix}-z \\x+y \\0\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

⇒ z = 0, x + y = 0

If we take x = 1 ⇒ y = –1

Hence, the corresponding eigen-vector X 1 = [1 –1 0] T .

Similarly, the eigenvector corresponding to 𝜆 = 2 is given by:

\(\begin{array}{l}\Rightarrow \begin{bmatrix}-1 & 0 & -1 \\1 & 0 & 1 \\2 & 0 & 1 \\\end{bmatrix}\begin{bmatrix}x \\y \\z\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

Applying elementary transformation R 3 → R 3 – R 2 and R 2 → R 2 + R 1 , we get

\(\begin{array}{l}\Rightarrow \begin{bmatrix}-1 & 0 & -1 \\0 & 0 & 0 \\1 & 2 & 0 \\\end{bmatrix}\begin{bmatrix}x \\y \\z\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

\(\begin{array}{l}\Rightarrow \begin{bmatrix}-x-z \\0 \\x+2y\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

⇒ x + z = 0, x + 2y = 0

⇒ x = –2, y = 1 and z = 2 {taking y = 1}

Hence, the corresponding eigen-vector X 2 = [ –2 1 2] T .

Finally, the eigenvector corresponding to 𝜆 = 3 is the non-zero solution of the following matrix equation:

\(\begin{array}{l}\Rightarrow \begin{bmatrix}-2 & 0 & -1 \\1 & -1 & 1 \\2 & 2 & 0 \\\end{bmatrix}\begin{bmatrix}x \\y \\z\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

Applying elementary transformation R 2 → R 2 + R 1 and R 3 → R 3 + 2R 2 , we get

\(\begin{array}{l}\Rightarrow \begin{bmatrix}-2 & 0 & -1 \\-1 & -1 & 0 \\0 & 0 & 0 \\\end{bmatrix}\begin{bmatrix}x \\y \\z\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

⇒ 2x + z = 0, x + y = 0

On taking x = 1, we get x = 1, y = –1 and z = –2

Hence, the corresponding eigen-vector X 3 = [ 1 –1 –2] T .

Let us construct a matrix with these eigenvectors as its column vectors, we get

\(\begin{array}{l}P=\begin{bmatrix}1 & -2 & 1 \\-1 & 1 & -1 \\0 & 2 & -2 \\\end{bmatrix}\end{array} \)

Inverse of P is

\(\begin{array}{l}P^{-1}=\frac{1}{2}\begin{bmatrix}0 & -2 & 1 \\-2 & -2 & 0 \\-2 & -2 & -1 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}P^{-1}AP=\frac{1}{2}\begin{bmatrix}0 & -2 & 1 \\-2 & -2 & 0 \\-2 & -2 & -1 \\\end{bmatrix}\begin{bmatrix}1 & 0 & -1 \\1 & 2 & 1 \\2 & 2 & 3 \\\end{bmatrix}\begin{bmatrix}1 & -2 & 1 \\-1 & 1 & -1 \\0 & 2 & -2 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}=\begin{bmatrix}1 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & 3 \\\end{bmatrix}\end{array} \)

= diag(1, 2, 3)

Thus, A is diagonalizable.

Refer: Diagonalization

Question 8:

Show that the transformation T: V 2 ( R ) → V 2 ( R ) defined by T(a, b) = (a + b, a) ∀ a, b ∈ R is a linear transformation.

To show that T is a linear transformation, we need to prove that,

For any x, y ∈ V 2 ( R )

T( x + y ) = T( x ) + T( y ) and T(a x ) = aT( x ) where a is a scalar in field.

Let (x 1 , y 1 ) and (x 2 , y 2 ) are arbitrary elements of V 2 ( R )

T[(x 1 , y 1 ) + (x 2 , y 2 )] = T[(x 1 + x 2 , y 1 + y 2 )] = (x 1 + x 2 + y 1 + y 2 , x 1 + x 2 ) …..(i)

T(x 1 , y 1 ) + T(x 2 , y 2 ) = (x 1 + y 1 , x 1 ) + (x 2 + y 2 , x 2 ) = (x 1 + x 2 + y 1 + y 2 , x 1 + x 2 ) …..(ii)

From (i) and (ii), we get T[(x 1 , y 1 ) + (x 2 , y 2 )] = T(x 1 , y 1 ) + T(x 2 , y 2 )

Now, T[a(x 1 , y 1 )] = T(ax 1 , ay 2 ) = (ax 1 + ay 1 , ax 1 ) = a(x 1 + y 1 , x 1 ) = aT(x 1 , y 1 ).

∴ T is a linear transformation.

Question 9:

Show that the given subset of vectors of R 3 forms a basis for R 3 .

{(1, 2, 1), (2, 1, 0), (1, –1, 2)}

S = {(1, 2, 1), (2, 1, 0), (1, –1, 2)}

We know that any set of n linearly independent vectors forms the basis of n-dimensional vector space.

Now, dim R 3 = 3, we just need to prove that vectors in S are linearly independent.

Let \(\begin{array}{l}A=\begin{bmatrix}1 & 2 & 1 \\2 & 1 & -1 \\1 & 0 & 2 \\\end{bmatrix}\end{array} \)

We reduce this matrix to row echelon form to check the rank of A.

Applying R 2 → R 2 + (–2)R 1 and R 3 → R 3 + ( –1)R 1 , we get

\(\begin{array}{l}A\sim\begin{bmatrix}1 & 2 & 1 \\0 & -3 & -3 \\0 & -2 & 1 \\\end{bmatrix}\end{array} \)

Applying R 2 → ( –⅓)R 2 and R 3 → R 3 + 2R 2 , we get

\(\begin{array}{l}A\sim\begin{bmatrix}1 & 2 & 1 \\0 & 1 & 1 \\0 & 0 & 3 \\\end{bmatrix}\end{array} \)

Clearly, rank of A = 3 = number of vectors.

Thus, the given vectors are linearly independent.

⇒ S forms the basis of R 3 .

Question 10:

Given a linear transformation T on V 3 ( R ) defined by T(a, b, c) = (2b + c, a – 4b, 3a) corresponding to the basis B = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. Find the matrix representation of T.

Now, T(1, 0, 0) = (2 × 0 + 0, 1 – 4 × 0, 3 × 1) = (0, 1, 3)

= 0(1, 0, 0) + 1(0, 1, 0) + 3(0, 0, 1)

T(0, 1, 0) = (2 × 1 + 0, 0 – 4 × 1, 3 × 0) = (2, –4, 0)

= 2(1, 0, 0) –4(0, 1, 0) + 0(0, 0, 1)

And T(0, 0, 1) = (2 × 0 + 1, 0 – 4 × 0, 3 × 0) = (1, 0, 0)

= 1(1, 0, 0) + 0(0, 1, 0) + 0(0, 0, 1)

Then, the matrix representation of T with respect to the basis B is

\(\begin{array}{l}[T ; B] = \begin{bmatrix}0 & 2 & 1 \\1 & -4 & 0 \\3 & 0 & 0 \\\end{bmatrix}\end{array} \)

Practice Problems on Linear Algebra

1. Show that the following matrix is diagonalizable:

\(\begin{array}{l}A=\begin{bmatrix}8 & -8 & -2 \\4 & -3 & -2 \\3 & -4 & 1 \\\end{bmatrix}\end{array} \)

2. Show that the transformation T: V 3 ( R ) → V 2 ( R ) defined by T(a, b, c) = (b, c) ∀ a, b, c ∈ R is a linear transformation.

3. Show that the given subset of vectors of R 3 forms a basis for V 3 ( R ).

{(1, 0, –1), (1, 2, 1), (0, –3, 2)}.

4. Given a linear transformation T on V 3 ( R ) defined by T(a, b, c) = (2b + c, a – 4b, 3a) corresponding to the basis B = {(1, 1, 1), (1, 1, 0), (1, 0, 0)}. Find the matrix representation of T.

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Linear Algebra

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• Linear Algebra
• Solving Linear Systems Using Matrices
• Determinants
• Eigenvalues and Eigenvectors
• Kernel (Nullspace)
• Vector Space
• Cayley-Hamilton Theorem
• Row And Column Spaces
• Spectral Theorem
• Fundamental Subspaces
• Change of Basis
• Rank-Nullity Theorem
• Linear Transformations
• Linear Independence
• Jordan Canonical Form
• Affine transformations

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Problems in Mathematics

• Linear Combination and Linear Independence
• The expression $c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots+c_k\mathbf{v}_k$ is called a linear combination of vectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\in \R^n$, where $c_1, c_2, \dots, c_k$ are scalars in $\R$.
• A set of vectors $\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\}$ is said to be linearly independent if the only scalrs $c_1, c_2, \dots, c_k$ satisfying $c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots+c_k\mathbf{v}_k=\mathbf{0}$ are $c_1=c_2=\cdots=c_k=0$. We also say that the vectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ are linearly independent.
• If vectors are not linearly independent, they are linearly dependent .
• The set of $n$-dimensional vectors $\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\}$ are linearly dependent if $k > n$. (If there are more vectors than the dimension, then the vectors are linearly dependent.)
• Express the vector $\mathbf{b}=\begin{bmatrix} 2 \\ 13 \\ 6 \end{bmatrix}$ as a linear combination of the vectors \[\mathbf{v}_1=\begin{bmatrix} 1 \\ 5 \\ -1 \end{bmatrix}, \mathbf{v}_2= \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \mathbf{v}_3= \begin{bmatrix} 1 \\ 4 \\ 3 \end{bmatrix}.\] ( The Ohio State University, Linear Algebra Exam )
• Write the vector $\begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix}$ as a linear combination of the vectors $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ , $\begin{bmatrix} 2 \\ -2 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 0 \\ 4 \end{bmatrix}$.
• For what value(s) of $a$ is the following set $S$ linearly dependent? \[ S=\left \{\,\begin{bmatrix} 1 \\ 2 \\ 3 \\ a \end{bmatrix}, \begin{bmatrix} a \\ 0 \\ -1 \\ 2 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ a^2 \\ 7 \end{bmatrix}, \begin{bmatrix} 1 \\ a \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ -2 \\ 3 \\ a^3 \end{bmatrix} \, \right\}.\] See (a)
• Prove that any set of vectors which contains the zero vector is linearly dependent.
• Let $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ be a set of nonzero vectors in $\R^m$ such that the dot product $mathbf{v}_i\cdot \mathbf{v}_j=0$ when $i\neq j$. Prove that the set is linearly independent. See (b)
• Determine whether the following set of vectors is linearly independent or linearly dependent. If the set is linearly dependent, express one vector in the set as a linear combination of the others. \[\left\{\, \begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}, \begin{bmatrix} -1 \\ -2 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} -2 \\ -2 \\ 7 \\ 11 \end{bmatrix}\, \right\}.\]
• Find the value(s) of $h$ for which the following set of vectors \[\left \{ \mathbf{v}_1=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \mathbf{v}_2=\begin{bmatrix} h \\ 1 \\ -h \end{bmatrix}, \mathbf{v}_3=\begin{bmatrix} 1 \\ 2h \\ 3h+1 \end{bmatrix}\right\}\] is linearly independent. ( Boston College )
• Let \[\mathbf{v}_1=\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \mathbf{v}_2=\begin{bmatrix} 1 \\ a \\ 5 \end{bmatrix}, \mathbf{v}_3=\begin{bmatrix} 0 \\ 4 \\ b \end{bmatrix}\] be vectors in $\R^3$. Determine a condition on the scalars $a, b$ so that the set of vectors $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is linearly dependent.
• Determine conditions on the scalars $a, b$ so that the following set $S$ of vectors is linearly dependent. \begin{align*} S=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}, \end{align*} where \[\mathbf{v}_1=\begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix}, \mathbf{v}_2=\begin{bmatrix} 1 \\ a \\ 4 \end{bmatrix}, \mathbf{v}_3=\begin{bmatrix} 0 \\ 2 \\ b \end{bmatrix}.\]
• Let $A$ be a $3\times 3$ matrix and let $\mathbf{v}=\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$ and $\mathbf{w}=\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$. Suppose that $A\mathbf{v}=-\mathbf{v}$ and $A\mathbf{w}=2\mathbf{w}$. Then find the vector $A^5\begin{bmatrix} -1 \\ 8 \\ -9 \end{bmatrix}$.
• (a) Prove that the column vectors of every $3\times 5$ matrix $A$ are linearly dependent. (b) Prove that the row vectors of every $5\times 3$ matrix $B$ are linearly dependent.
• Suppose $M$ is an $n \times n$ upper-triangular matrix. If the diagonal entries of $M$ are all non-zero, then prove that the column vectors are linearly independent. Does the conclusion hold if we do not assume that $M$ has non-zero diagonal entries?
• Suppose that an $n \times m$ matrix $M$ is composed of the column vectors $\mathbf{b}_1 , \cdots , \mathbf{b}_m$. Prove that a vector $\mathbf{v} \in \R^n$ can be written as a linear combination of the column vectors if and only if there is a vector $\mathbf{x}$ which solves the equation $M \mathbf{x} = \mathbf{v}$.
• Introduction to Matrices
• Elementary Row Operations
• Gaussian-Jordan Elimination
• Solutions of Systems of Linear Equations
• Nonsingular Matrices
• Inverse Matrices
• Subspaces in $\R^n$
• Bases and Dimension of Subspaces in $\R^n$
• General Vector Spaces
• Subspaces in General Vector Spaces
• Linearly Independency of General Vectors
• Bases and Coordinate Vectors
• Dimensions of General Vector Spaces
• Linear Transformation from $\R^n$ to $\R^m$
• Linear Transformation Between Vector Spaces
• Orthogonal Bases
• Determinants of Matrices
• Computations of Determinants
• Introduction to Eigenvalues and Eigenvectors
• Eigenvectors and Eigenspaces
• Diagonalization of Matrices
• The Cayley-Hamilton Theorem
• Dot Products and Length of Vectors
• Eigenvalues and Eigenvectors of Linear Transformations
• Jordan Canonical Form

Free Mathematics Tutorials

Eigenvalues and eigenvectors questions with solutions.

Examples and questions on the eigenvalues and eigenvectors of square matrices along with their solutions are presented. The properties of the eigenvalues and their corresponding eigenvectors are also discussed and used in solving questions.

Page Content

Definition of eigenvalues and eigenvectors, examples with solutions on eigenvalues and eigenvectors, eigenvalues of triangular matrices, eigenvalues of the power of a matrix, properties of eigenvalues and eigenvectors.

• Solutions to the Questions

Let A be an n × n square matrix. If there exist a non trivial (not all zeroes) column vector X solution to the matrix equation A X = λ X ; where λ is a scalar, then X is called the eigenvector of matrix A and the corresponding value of λ is called the eigenvalue of matrix A. Let us rewrite the matrix equation in standard form: A X - λ X = 0 Let I be the n × n identity matrix and substitute X by I X in the above equation A X - λ I X = 0 Rewrite as (A - λ I) X = 0 The above matrix equation has non trivial solutions if and only if the determinant of the matrix (A - λ I) is equal to zero. Det (A - λ I) = 0 is called the characteristic equation of A. If A is an n by n matrix, when (A - λ I) is expanded, it is a polynomial of degree n and therefore (A - λ I) is called the characteristic polynomial of A.

Example 2 Find all eigenvalues and eigenvectors of matrix \[ A = \begin{bmatrix} 1 & 0 & -1 \\ 1 & 0 & 0 \\ -2 & 2 & 1 \end{bmatrix} \] Solution Find Eigenvalues We first find the matrix \( A - \lambda I \). \( A - \lambda I = \begin{bmatrix} 1 & 0 & -1 \\ 1 & 0 & 0 \\ -2 & 2 & 1 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0& 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 - \lambda & 0 & -1 \\ 1 & - \lambda & 0 \\ -2 & 2 & 1 - \lambda \end{bmatrix}\) Write the characteristic equation. \( Det(A - \lambda I) = (1-\lambda)(-\lambda(1-\lambda)) - 1(2 - 2\lambda) = 0 \) factor and rewrite the equation as \( (1 - \lambda)(\lambda - 2)(\lambda+1) = 0 \) which gives 3 solutions \( \lambda = - 1 , \lambda = 1 , \lambda = 2 \) Find Eigenvectors Eigenvectors for \( \lambda = - 1 \) Substitute \( \lambda \) by - 1 in the matrix equation \( (A - \lambda I) X = 0 \) with \( X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \) \( \begin{bmatrix} 2 & 0 & -1 \\ 1 & 1 & 0 \\ -2 & 2 & 2 \end{bmatrix} \ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \) Row reduce to echelon form gives \( \begin{bmatrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 0 \end{bmatrix} \ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \) The solutions to the above system and are given by \( x_3 = t , x_2 = -t/2 , x_1 = t/2, t \in \mathbb{R} \) Hence the eigenvector corresponding to the eigenvalue \( \lambda = -2 \) is given by \( X = t \begin{bmatrix} 1/2 \\ -1/2 \\ 1 \end{bmatrix} \) Eigenvectors for \( \lambda = 1 \) Substitute \( \lambda \) by \( 1 \) in the matrix equation \( (A - \lambda I) X = 0 \). \( \begin{bmatrix} 0 & 0 & -1 \\ 1 & - 1 & 0 \\ -2 & 2 & 0 \end{bmatrix} \ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \) Row reduce to echelon form gives \( \begin{bmatrix} 1 & - 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \) The solutions to the above system and are given by \( x_3 = 0 , x_2 = t , x_1 = t , t \in \mathbb{R} \) Hence the eigenvector corresponding to the eigenvalue \( \lambda = 1 \) is given by \( X = t \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \) Eigenvectors for \( \lambda = 2 \) Substitute \( \lambda \) by \( 2 \) in the matrix equation \( (A - \lambda I) X = 0 \). \( \begin{bmatrix} -1 & 0 & -1 \\ 0& - 2 & 0 \\ -2 & 2 & -1 \end{bmatrix} \ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \) Row reduce to echelon form gives \( \begin{bmatrix} 1 & 0 & 1 \\ 0 & -2 & -1 \\ 0 & 0 & 0 \end{bmatrix} \ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \) The solutions to the above system and are given by \( x_3 = t , x_2 = -t/2 , x_1 = - t , t \in \mathbb{R} \) Hence the eigenvector corresponding to the eigenvalue \( \lambda = 2 \) is given by \( X = t \begin{bmatrix} -1 \\ -1/2 \\ 1 \end{bmatrix} \)

If \( \lambda \) is an eigenvalue of matrix A, then we can write \( AX = \lambda X \), where X is the eigenvector corresponding to the eigenvalue \( \lambda \). Left multiply both sides of the above equation by matrix A. \( A (A X) = A (\lambda X) \) Simplify to \( A^2 X = A (\lambda X) = \lambda (A X)\) Substitute \( A X \) on the right side by \( \lambda X \) to obtain \( A^2 X = \lambda^2 X\) We can continue multiplying by A and simplifying to obtain \( A^n X = \lambda^n X\) If \( \lambda \) is an eigenvalue of matrix A and X the corresponding eigenvector, then \( \lambda^n \) is an eigenvalue of matrix \( A ^n\) and X its corresponding eigenvector. Videos at Find Eigenvectors and Eigenvalues of a 2 by 2 Matrix on Video and Find Eigenvectors and Eigenvalues of a 3 by 3 Matrix on Video

• Matrix A is singular if and only if \( \lambda = 0 \) is an eigenvalue value of matrix A. or If matrix A is invertible, then none of its eigenvalues is equal to zero.
• If \( \lambda \) is an eigenvalue of matrix A and X the corresponding eigenvector, then the eigenvalue of matrix \( A ^n\) is equal to \( \lambda^n \) and the corresponding eigenvector is X.
• The product of all the eigenvalues of a matrix is equal to its determinant.
• The sum of all the eigenvalues of a matrix is equal to its trace (the sum of all entries in the main diagonal). You may check the examples above.
• The eigenvalues of matrix A and its transpose are the same.
• If A is a square invertible matrix with \( \lambda \) its eigenvalue and X its corresponding eigenvector, then \( 1/\lambda \) is an eigenvalue of \( A^{-1} \) and X is a corresponding eigenvector.
• If \( \lambda \) is an eigenvalue of matrix A and X a corresponding eigenvalue, then \( \lambda - t \) , where t is a scalar, is an eigenvalue of \( A - t I \) and X is a corresponding eigenvector.
• Part 1 1) Find all eigenvalues and their corresponding eigenvectors for the matrices: a) \( A = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} \) , b) \( B = \begin{bmatrix} -1 & 1 & 1 \\ 2 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} \) Part 2 1) Find all values of parameters p and q for which the matrix \( A = \begin{bmatrix} 2 & p\\ 2 & q \end{bmatrix} \) has eigenvalues equal to - 1 and -3. 2) Find all values of parameters p which the matrix \( A = \begin{bmatrix} 1 & -33 & -1\\ 0 & p - 1 & 3 \\ 0 & 0 & p + 1 \end{bmatrix} \) has eigenvalues equal to 1 and 2 and 3.
• Part 3 Matrix \( A = \begin{bmatrix} 1 & - 1\\ 2 & p \end{bmatrix} \) , where p is a parameter, has an eigenvector \( X = \begin{bmatrix} 1\\ -1 \end{bmatrix} \). Find all the eigenvalues and eigenvectors of matrix A.
• Part 4 Matrix \( A = \begin{bmatrix} a & b\\ 1 & -1 \end{bmatrix} \) has eigenvalues 3 and 4. Find the eigenvectors of matrix A.
• Part 5 A 3 by 3 matrix A has eigenvalues 1 ,2 and 3 and their corresponding eigenvectors \( \begin{bmatrix} 1\\ 0 \\ 1 \end{bmatrix} \), \( \begin{bmatrix} 0\\ -1 \\ 1 \end{bmatrix} \) and \( \begin{bmatrix} 1/2\\ 1 \\ 0 \end{bmatrix} \). Find the product \( A \begin{bmatrix} 3\\ -1 \\ 2 \end{bmatrix} \).
• Part 6 Matrix \( A \) has eigenvalues 2 and 3 and their corresponding eigenvectors \( \begin{bmatrix} -1\\ 1 \end{bmatrix} \) and \( \begin{bmatrix} -1/2\\ 1 \end{bmatrix} \). Find the eigenvalues and the corresponding eigenvectors of \( A^{-3} \).

Solutions to the Above Questions

• Part 1 Matrix A \( A - \lambda I = \begin{bmatrix} 1 - \lambda & 0\\ -1 & 2-\lambda \end{bmatrix} \) Characteristic equation \( (1-\lambda)(2-\lambda) = 0 \) Eigenvalues are solutions to the above equation; there are two solutions. \( \lambda = 1 \) and \( \lambda = 2 \)
• Eigenvectors for \( \lambda = 1 \) \( A - \lambda I = \begin{bmatrix} 0 & 0\\ -1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0\) Eigenvector is the solution to the above system which can be written as \( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = t \begin{bmatrix} 1 \\ 1 \end{bmatrix} , t \in \mathbb{R} \) Eigenvectors for \( \lambda = 2 \) \( A - \lambda I = \begin{bmatrix} -1 & 0\\ -1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0\) Eigenvector is the solution to the above system which can be written as \( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = t \begin{bmatrix} 0 \\ 1 \end{bmatrix} , t \in \mathbb{R} \) Matrix B \( B - \lambda I = \begin{bmatrix} -1 - \lambda & 1 & 1\\ 2 & 1 -\lambda & 1 \\ 1 & 0 & -\lambda \end{bmatrix} \) Using the last row to find determinant, the characteristic equation \( - \lambda^3 + 4\lambda = 0 \) Eigenvalues are solutions to the above equation and are given by: \( \lambda = 0 \) , \( \lambda = 2 \) and \( \lambda = - 2 \) Eigenvectors for \( \lambda = 0 \) \( B - \lambda I = \begin{bmatrix} -1 & 1 & 1\\ 2 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \) Rewrite in row echelon form \( \begin{bmatrix} -1 & 1 & 1\\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0\) Eigenvector is the solution to the above system which can be written as \( \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = t \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} , t \in \mathbb{R} \) Eigenvectors for \( \lambda = 2 \) \( B - \lambda I = \begin{bmatrix} -3 & 1 & 1\\ 2 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0\) Rewrite in row echelon form \( \begin{bmatrix} 1 & 0 & - 2\\ 0 & - 1 & 5 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0\) Eigenvector is the solution to the above system which can be written as \( \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = t \begin{bmatrix} 2 \\ 5 \\ 1 \end{bmatrix} , t \in \mathbb{R} \) Eigenvectors for \( \lambda = - 2 \) \( B - \lambda I = \begin{bmatrix} 1 & 1 & 1\\ 2 & 3 & 1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0\) Rewrite in row echelon form \( \begin{bmatrix} 1 & 1 & 1\\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0\) Eigenvector is the solution to the above system which can be written as \( \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = t \begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix} , t \in \mathbb{R} \)
• Part 2 \( A - \lambda I = \begin{bmatrix} 2 - \lambda & p\\ 2 & q -\lambda \end{bmatrix} \) The characteristic equation is given by \( (2 - \lambda)(q - \lambda ) - 2p = 0 \) The eigenvalues are given as - 1 and -3 and are solutions to the characteristic equation. Substitute \( \lambda \) by - 1 and -3 to obtain a system of equations in p and q. \( 3(q + 1 ) - 2p = 0 \) and \( 5(q + 3 ) - 2p = 0 \) Solve to obtain p = -15/2 and q = -6.
• Part 3 Let \( \lambda \) be the eigenvalue corresponding to the given eigenvector. Hence \( \begin{bmatrix} 1 & - 1\\ 2 & p \end{bmatrix} \begin{bmatrix} 1\\ -1 \end{bmatrix} = \lambda \begin{bmatrix} 1\\ -1 \end{bmatrix} \) Use the above matrix equation to write a system of equations in p and \( \lambda \) as follows: \( 1 + 1 = \lambda \) and \( 2 - p = - \lambda \) Solve to obtain p = 4 and \( \lambda = 2 \) We can write matrix A as \( A = \begin{bmatrix} 1 & - 1\\ 2 & 4 \end{bmatrix} \) The product of the eigenvalues is equal to the determinant of A (property 3 above). Hence \( Det (A) = 4 + 2 = 2 \lambda \) gives the second eigenvalue as \( \lambda = 3 \) The eigenvector corresponding to \( \lambda = 3 \) is given by \( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = t \begin{bmatrix} -1/2 \\ 1 \end{bmatrix} , t \in \mathbb{R} \)
• Part 4 The product of the eigenvalues is equal to the determinant of A. Hence \( Det (A) = - a - b = 3 (4) = 12 \) The sum of the eigenvalues is equal to the trace. Hence a - 1 = 3 + 4 = 7 Solve the two equations in a and b simultaneously to find a = 8 and b = -20 Hence matrix A is given by \( A = \begin{bmatrix} 8 & -20\\ 1 & -1 \end{bmatrix} \) and its eigenvectors of are given by: \( \begin{bmatrix} 4\\ 1 \end{bmatrix} \) and \( \begin{bmatrix} 5\\ 1 \end{bmatrix} \).
• Part 5 Combine the three eigenvalues and eigenvectors to write \( A \begin{bmatrix} 1 & 0 & 1/2\\ 0 & -1 & 1\\ 1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 \begin{bmatrix} 1\\ 0 \\ 1 \end{bmatrix} & 2 \begin{bmatrix} 0\\ -1 \\ 1 \end{bmatrix} & 3\begin{bmatrix} 1/2\\ 1 \\ 0 \end{bmatrix} \end{bmatrix} \) Hence \( A = \begin{bmatrix} 1 & 0 & 3/2\\ 0 & -2 & 3 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1/2\\ 0 & -1 & 1\\ 1 & 1 & 0 \end{bmatrix}^{-1}\) which then gives \( A \begin{bmatrix} 3\\ -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 3/2\\ 0 & -2 & 3 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1/2\\ 0 & -1 & 1\\ 1 & 1 & 0 \end{bmatrix}^{-1} \begin{bmatrix} 3\\ -1 \\ 2 \end{bmatrix} = \begin{bmatrix}-1\\ -6\\ -1\end{bmatrix}\)
• Part 6 According to property 6 above, the eigenvalues of \( A^{-1} \) are 1/2 and 1/3 and the corresponding eigenvectors are \( \begin{bmatrix} -1\\ 1 \end{bmatrix} \) and \( \begin{bmatrix} -1/2\\ 1 \end{bmatrix} \). By definition \( A^{-3} \) = \( (A^{-1})^3 \). Hence according to property 2 above, the eigenvalues of \( A^{-3} \) are \( (1/2)^3 = 1/8 \) and \( (1/3)^3 = 1/27 \) and the corresponding eigenvalues are \( \begin{bmatrix} -1\\ 1 \end{bmatrix} \) and \( \begin{bmatrix} -1/2\\ 1 \end{bmatrix} \).

More References and links

• Find Eigenvectors and Eigenvalues of a 2 by 2 Matrix on Video
• Matrices with Examples and Questions with Solutions .
• Determinant of a Square Matrix .
• Inverse Matrix Questions with Solutions .
• Determinant Howard Anton, Chris Rorres - Elementary Linear Algebra - ISBN 0-471-58741-9 - 7 th Edition

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