Test by Bikram | Mock GATE | Test 4 | Question: 25

Question

The Eigen values of a $2 \times 2$ matrix $’A’$ are $1, -2$, and its Eigen vectors $x_1$ and $x_2$ respectively.

The Eigen values and Eigen vectors of the matrix $A^{2} - 3A + 4I$  (where $I$ is the identity matrix) will be:

• A.   $2,14$ and $ x_1,x_2$
• B.   $2,14$ and $x_1+x_2 , x_1- x_2$
• C.   $2,0$ and $x_1,x_2$
• D.   $2,0$ and $x_1+x_2, x_1- x_2$

Answer 1

if for matrix $A, m$ is eigen value and x is corresponding eigen vector then $Ax = mx$

  $Ax$ $=$ $mx$ $\Rightarrow$ $\left (A^{2} \right )$$\times$ $=$ $\left (m^{2} \right )$$\times$

  $Ax$ = $mx$ $\Rightarrow$ $\left ( nA \right )$$\times$ $=$  $\left ( nm \right )$$\times$for any real $n$

Now, finding ($a^{2}$ - $3$$A$ + $4$$I$$\times 1$ and $a^{2}$ - $3$$A$ + $4$$I$$\times 2$ will not be difficult.

Answer will be $(A)$.

Comments

So, you are saying eigen vectors never change due to every expression i can do like this @Ashwani Kumar 2

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Yes what I thought...it seems correct according to expression $AX=\lambda X$

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http://thejuniverse.org/PUBLIC/LinearAlgebra/MATH-232/Unit.12/Presentation.2/Section12A/properties.html#:~:text=If%20a%20matrix%20A%20has,to%20the%20same%20eigenvector%20x.&text=Suppose%20the%20matrix%20A%20has,x)%20%3D%20%CE%BB2x.