CSIR NET Solved problems on Eigenvalue and Eigenvector.

1..Suppose the matrix$A =\begin{pmatrix}40&-29&-11\\-18&30&-12\\26&24&-50\end{pmatrix}$

has a certain complex number $\lambda \neq 0$ as an eigenvalue. 

Which of the following numbers must also be an eigenvalue of A ?

• $\lambda + 20$
• $\lambda - 20$
• $20 - \lambda$
• $- 20 - \lambda$

Solution:

$A = \begin{pmatrix} 40 &-29&-11\\-18 & 30&-12\\ 26 & 24&-50\end{pmatrix}\rightarrow\begin{pmatrix} 0&-29&-11\\0 & 30&-12\\0 & 24&-50\end{pmatrix}$ $(C_1 \to C_1+C_2+C_3)$

So, clearly determinant of this matrix is 0. So. One eigenvalue=$0$.

And since Trance=$20$ So, third eigenvalue=$20-\lambda$

So, 3rd option is correct.

2.Let $A$ be a $3\times 3$ matrix with real entries such that $\det(A)=6$ and the trace of $A$ is $0$ . If $\det(A+I)=0$ , where $I$ denote the 3\times 3 identity matrix, then the eigenvalues of $A$ are

• $-1,2,3$

• $-1,2,-3$

• $1,2,-3$

• $-1,-2,3$

Solution:

Some Results Related to Eigenvalue often used: If $\lambda$ is an eigenvalue of A then

o $k\lambda$ is an eigenvalue of $kA$

o $\lambda+k$ is an eigenvalue of $A+k\lambda I$

o $\lambda^2$ is an eigenvalue of $A^2$

o $\frac{|A|}{\lambda}$ is an eigenvalue of $adj(A)$ provided $|A|\neq 0$

o $\frac{1}{\lambda}$ is an eigenvalue of $\frac{1}{A}$ provided $|A|\neq 0$

o $P(\lambda)$ is an eigenvalue of $P(A)$

Also,

Sum of eigenvalues=Trace of matrix product of eigenvalues=Product of eigenvalues.

So, Let, $a,b,c$ are eigenvalues of a then we have $a+b+c=0$ and $abc=6$

And since determinant of $A+I=0$ therefore, It’s one eigenvalue is $0$ So, one eigenvalue of $A$ Must be equal to $-1$.

So, we have $a+b-1=0$ and $ab=-6$ Solving we get, eigenvalues of $A=-1,-2,3$.

So, 4th option is correct.

3.Let $P$ be a $2 \times 2$  complex matrix such that $P^* P$  is the identity matrix, where $P^*$ is the conjugate transpose of $P$ . Then the eigenvalue of $P$ are

• real

• complex conjugate of each other

• reciprocals of each other

• of modulus $1$

Solution:

Some basis Results:

Eigenvalues of Symmetric and Hermitian Matrix are Real

Eigenvalues of Skew Symmetric and Skew-Hermitian are Imaginary.

Modulus of Eigenvalues of Unitary and orthogonal Matrix =|z|=1

Eigenvalue of Nilpotent Matrix=0.

You can remember these results using this diagram.

So, using above result, 4th option is correct.

4.Let $\{v_1,\cdots , v_n\}$ be a linearly independent subset of a vector space $V$ where $n \geq 4$ . Set $w_{ij} =v_i-v_j$ . Let $W$ be the span of $\{w_{ij}|1\leq i,j\leq n\}$ . Then

•$\{w_{ij}|1\leq i$

• $\{w_{ij}|1\leq i$

• $\{w_{ij}|1\leq i\leq n-1,j=i+1\} spans W$ .

• $\dim W=n$

Solution: 

Let $S=\{w_{ij}|1\leq i\leq n-1,j=i+1\} =\{w_{12},w_{23},\cdots,w_{n-1 n}\}$

We will prove that $S$ set spans $W$.

Let $w_{mn} \in W$ Let $m>n$ then

$w_{mn}$

=$v_m-v_n$

=$-((v_n-v_{n+1})+(v_{n+1}-v_{n+2})\cdots (v_{m-1}-v_m))$

=$-(w_{n,n+1}+w_{n+1,n+2}\cdots w_{m-1,m})$

 If $n>m$ then

$w_{mn}$

$=v_m-v_n$

$=(v_m-v_{m+1})+(v_{m+1}-v_{m+2})\cdots (v_{n-1}-v_n)$

$=w_{m,m+1}+w_{m+1,m+2}\cdots w_{n-1,n}$

So, $\{w_{ij}|1\leq i\leq n-1,j=i+1\}$ spans $W$ .

Also since it contains only $n-1$ terms. so, dimension of $W=n-1$.

(Since the basis set is always, the subset of spanning set)

Also, $\{w_{ij}|1\leq i$

$\{w_{ij}|1\leq i\leq n-1,j=i+1\}$

Therefore it will also span $W$

So, 1,3 are correct.

5.Which of the following matrices is not diagonalizable over $\mathbb{R}$ ?

• $\begin{pmatrix} 1&1&0\\ 0&2&0\\0&0&2\end{pmatrix}$

• $\begin{pmatrix} 1&1&0\\0&2&1\\0&0&3 \end{pmatrix}$

• $\begin{pmatrix} 1&1&0\\0&1 &0\\0&0&2\end{pmatrix}$

• $\begin{pmatrix} 1&0&1\\0&2&0\\0&0&3\end{pmatrix}$

Solution: 

We don't need to find eigenvalues.

Notice that

$\begin{pmatrix} 1&1&0\\ 0&1 &0\\ 0&0&2 \end{pmatrix}$

 is in Standard Jordan Canonical form. 

Hence 3rd option is correct.

6.Let $A$ be a $3\times 3$ matrix with $A^3=-1$ . Which of the following statements are correct?

• $A$ has three distinct eigenvalues.

• $A$ is diagonalizable over $\mathbb{C}$ .

• $A$ is triangularizable over $\mathbb{C}$ .

• $A$ is non singular.

Solution:

 $A^3=-1$

Therfore it's we have characteristic polynomial

$\lambda^3=-1$

$\Rightarrow \lambda^3+1=0$

$\Rightarrow (\lambda+1)(\lambda^2-2\lambda+1=0$

$\Rightarrow (\lambda+1)(\lambda-1)^2$ so, eigenvalues are $1,1,-1$

Since they are not distinct, we can't say that It is diagonizable or not.

We know that every matrix is triangularizable.

and determinant of matrix is product of eigenvalues.

So. $|A|=-1$

So, 3rd and 4th options are correct.

7.Let $N$ be a non-zero $3\times 3$ matrix with the property $N^2=0$ . Which of the following is/are true?

• N is not similar to a diagonal matrix.

• N is similar to a diagonal matrix.

• N has non-zero eigenvector.

• N has three linearly independent eigenvectors.

Solution:

$N$ is basically a Nilpotent matrix and we know that,

 Nilpotent matrix has all Eigenvalue equals to $0$.

and nilpotent matrix is only digonalizable if it is a Zero matrix.

So, let $N=\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}$

This matrix satisfies the above conditions and eigenvectors are $\begin{pmatrix}1\\0\\0\end{pmatrix}\quad \begin{pmatrix}0\\1\\0\end{pmatrix}.$

So, 1st and 3rd options are correct.

8.Let $A$ be a $3 \times 3$ and $b$ be a $3 \times 1$ matrix with integer entries. Suppose that the system $Ax = b$ has a complex solution. Then

• $Ax = b$ has an integer solution

• $Ax = b$ has an rational solution

• The set of real solutions to $Ax = 0$ has a basis consisting of rational solutions.

• If $b\neq 0$ then $A$ has positive rank.

Solution:

Option:1&2 $x=A^{-1}b=\frac{adj(A)}{|A|}\times b$ now since integers are closed under the operation $+,-,\times, /$ So, numerator and denominator are integers.

So, division of two integers is rational number.

So, $b$ is rational.

Option:3

If you have Complex solution to $Ax=b$ (imaginary part of complex solution is not zero) then it is sure that you have at least one non-zero solution to $Ax=0$.

To illustrate this:

 Let $A$ be a $3\times 4$ and $b$ be a $3\times 1$ matrix with integer entries

.Suppose that the system $Ax=b$ has a complex solution.

 Let this complex solution is $u=\left( \begin{array}{ccc}x_1+i\times y_1\\x_2+i\times y_2\\x_3+i\times y_3\end{array} \right)$

$= \left( \begin{array}{ccc}x_1\\x_2 \\x_3\end{array} \right)+i\left( \begin{array}{ccc}y_1\\y_2\\y_3\end{array}\right)$

$=X+i \times Y$ ( where Y is not a zero vector)

So this system of linear equation can be written as: $A(X+i \times Y)=b +0$  since entries of $b$ are integers..i.e., real so we can consider it as two system of linear equations: $A X=b$ and $AY=0$   that means $AX=0$ has a solution other than zero. (which is $Y$). So it is clear that null space has dimension greater than equals to $1$

Option 4: 4th option is true since if $A=0$ then there exists no such $x$ such that $Ax\neq 0$ So, It's true

So, 2,3,and 4 are correct.

9. Let $S:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be given by $v \longmapsto \alpha v$ for a fixed $\alpha \in \mathbb{R} , \alpha \neq 0$ . Let $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be a linear transformation such that $\mathbb{B}\{v_1,\cdots , v_n\}$ is a set linearly independent eigenvectors of $T$ . Then

• The matrix of $T$ with respect to $\mathbb{B}$ is diagonal.

• The matrix of $T-S$ with respect to $\mathbb{B}$ is diagonal.

• The matrix of $T$ with respect to $\mathbb{B}$ is not necessarily diagonal, but upper triangular.

• The matrix of $T$ with respect to $\mathbb{B}$ is diagonal but the matrix of  $(T-S)$ with respect to $\mathbb{B}$ is not diagonal.

Solution:

Given, $\{x_1,\cdots , x_n\},$ is basis of $\mathbb{R}$. Also this set is eigenvectors of $T$ then,

$T(x_1)=\lambda_1\times x_1+0+\cdots$

$T(x_2)=0+\lambda_2\times x_2+0+\cdots$

$T(x_3)=0+0+\lambda_3\times x_3+0+\cdots$

$\cdots \cdots$

So, Matrix of linear transformation of $T$ with respect to $B$.

is

$\begin{pmatrix}\lambda_1&0&0&0&\cdots&\cdots\\0&\lambda_2&0&0&\cdots&\cdots\\0&0&\lambda_3&0&\cdots&\cdots\\\cdots&\cdots&\cdots&\cdots\\\cdots&\cdots&\cdots&\cdots\end{pmatrix}$

So, clearly. It is a Diagonal Matrix.

Again

$S(x_1)=\alpha a_1\times x_1+0+\cdots$

$S(x_2)=0+\alpha\times x_2+0+\cdots$

$S(x_3)=0+0+\alpha\times x_3+0+\cdots$

$\cdots \cdots$

So, Matrix of linear transformation of $S$ with respect to $B$.

$\begin{pmatrix}\alpha&0&0&0&\cdots&\cdots\\0&\alpha&0&0&\cdots&\cdots\\0&0&\alpha&0&\cdots&\cdots\\\cdots&\cdots&\cdots&\cdots\\\cdots&\cdots&\cdots&\cdots\end{pmatrix}$

So, Matrix of S is also a diagonal Matrix.

and since, $(T-S)(x)=T(x)-S(x)$ So, the matrix of $T-S$ is also Diagonal.

1,2 options are correct.

10. For an $n \times n$ real matrix $A,\lambda \in \mathbb{R}$ and a nonzero vector $v\in \mathbb{R}^n$ suppose that $(A-\lambda I)^kv=0$ for some

positive integer k . Let I be the n \times n identity matrix. Then which of the following is/are always true?

• $(A-\lambda I)^{k+r}v=0$ for all positive integer $r$ .

• $(A-\lambda I)^{k-1}v=0$

• $(A-\lambda I)$ is not injective.

• $\lambda$ is an eigenvalue of $A$ .

Solution:

$(A-\lambda I)^kv=0$

If $(A-\lambda I)v\neq 0$ then $(A-\lambda I)^{k-1}v= 0$

Proceeding in similar manner we get $(A-\lambda I)v=0$ (Contradiction!)

Therefore, $\lambda$ is an eigenvalue of $A$.

1,2,4 options are correct.

11. Let A\in M_{10}(\mathbb{C}) , the vector space of 10\times 10 matrices with entries in \mathbb{C} . Let W_A be the subspace of M_{10}(\mathbb{C}) spanned by \{A^n|n \geq =0\} . Choose the correct statements.

• for any $A, \dim(W_A)\leq 10$

• for any $A, \dim(W_A)< 10$

• for some $A, 10<\dim(W_A)< 100$

• for any $A, \dim(W_A)= 100$

Solution:

For a $10\times 10$ matrix there exists a characteristic polynomial of degree $10$ satisfied by $A$.

Also , there may exists a polynomial of degree< characteristic polynomial satisfied by $A$ [known as Monic Polynomial]

Let the characteristic polynomial is: $a_0+a_1\lambda+a_2 \lambda^2+\cdots +a_{10} \lambda^{10}=0$

So, by Cayley-Hamilton Theorem:

$a_0+a_1\times A+a_2\times A^2+\cdots +a_{10} A^{10}=0$

So, $W_A$ be the subspace of spanned by $\{A^n|n \geq =0\}$ has atmost $10$ elements.

So, 1st option is correct.

• • • • • • • • • • •