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Let A be 4 X 4 matrix with eigenvalues -5, -2, 1, 4. Which of the following option is an eigenvalue of $\begin{bmatrix}A & I\\I & A\end{bmatrix}$, where I is 4 X 4 identity matrix?

options: (A) -5  (B) -7 (C) 2 (D) 1
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1 Answer

up vote 8 down vote accepted

Hint: Suppose that $x=[x_1,x_2,x_3,x_4]^T$ is an eigenvector of $A$ corresponding to an eigenvalue $\lambda$, and let $y=[x_1,x_2,x_3,x_4,x_1,x_2,x_3,x_4]^T$. What can you say, then, about $$\left[\begin{array}{cc}A&I\\I&A\end{array}\right]y?$$

Edit: It may be that your difficulty lies in the fact that you're thinking of it as a matrix of matrices in the first place. That isn't really what is going on, but if that's how you're thinking of it, then instead consider $y$ as a "vector of vectors". So, $x$ is an eigenvector of $A$ corresponding to one of its eigenvalues, $\lambda$, then $y=\left[\begin{array}{c}x\\x\end{array}\right]$. Now simply doing matrix multiplication, we have $$\left[\begin{array}{cc}A&I\\I&A\end{array}\right]y=\left[\begin{array}{cc}A&I\\I&A\end{array}\right]\left[\begin{array}{c}x\\x\end{array}\right]=\left[\begin{array}{c}Ax+Ix\\Ix+Ax\end{array}\right]=\left[\begin{array}{c}\lambda x+x\\x+\lambda x\end{array}\right]=(\lambda+1)\left[\begin{array}{c}x\\x\end{array}\right]=(\lambda+1)y.$$ Thus, whenever $\lambda$ is an eigenvalue of $A$, we automatically have $\lambda+1$ as an eigenvalue of $\left[\begin{array}{cc}A&I\\I&A\end{array}\right]$. Thus, only one of the options given is necessarily an eigenvalue of the matrix of matrices.

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Can I ask for a little more hint? –  Bharat Kul Ratan Oct 6 '12 at 0:00
    
Does that get you the rest of the way? I couldn't really think of anything else to say that wasn't going to just give you the answer. :P –  Cameron Buie Oct 6 '12 at 1:04
    
yes, you were right. I was thinking this as a matrix of matrix and so couldn't figure out hint. thanks :) –  Bharat Kul Ratan Oct 6 '12 at 3:17
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