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I am given 4 choices for what can be deduced from the eigenvalues of matrix $B$, and 3 of them are correct and I have to choose which ones. The options are:

a) The rank or matrix $B$

b) The determinant of $B^TB$

c) The eigenvalues of $B^TB$

d) The eigenvalues of $(B^2+I)^-$$^1$

Okay so I think a) is true because apparently if one of the eigenvalues is 0 then the rank can be deduced.

The determinant I assume you can but I don't remember how.

I think d) is possible to calculate too but c) is not.

I just don't really know how to systematically think about these and would appreciate some explanations.

mangopancake
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1 Answers1

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You can definitely find a and b, since having eigenvalue 0 means that B is singular, and thus has a determinant of zero, then $det(B^T B)=det(B^T)det(B)=0$, so that is b done. You have your own method for a.

Now we are left with c and d, now d cannot hold since $(B ^2+I)^{-1}=\sum_{k=0}^\infty (-1)^kB^{2k}$ which only exists if all the eigen values of $B^2$ are smaller than one, but one can show that the eigenvalues of $B^2$ are 0,1 and 4. So it does not exist so you can't find the eigenvalues.

So it must be that a b and c hold.

Ellya
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  • I disagree with your claim that "d cannot hold". $2I$ has eigen values that are greater than one, but it's inverse certainly exists. You can show that $B^2 + I$ has eignevalues $1, 2, 5$, and hence its inverse has eigenvalues $ 1, \frac{1}{2}, \frac{1}{5} $. – Calvin Lin Mar 30 '14 at 18:45
  • No that's not what I'm saying, the sum is analogous to the geometric sum, which is why we need the eigenvalues to be less than one, note that I never mentioned what the eigenvalues of $B ^2+I $ were. – Ellya Mar 30 '14 at 18:48
  • To be precise about what I mean, take a step back and think of $\mathbb{R}$, here $\frac{1}{a^2+1}=(a^2+1)^{-1}=\sum_{k=0}^{\infty}(-1)^ka^{2k}$ the sum converges iff $|a^2|\lt 1$, this crosses over anologously to matrices, but we need eigenvalues less than one instead. – Ellya Mar 30 '14 at 18:56
  • Why is it that it crosses over analogously to matrices with eigenvalues instead? Also, for 2 x 2 matrices the eigenvalues of $B+I$ are the same as those of $B$ with 1 added to each. Does this also work for larger matrices? – mangopancake Mar 30 '14 at 19:10
  • I don't believe that holds in general no. But this stuff pops up in operator theory, and is very useful, so it is good to know these series. – Ellya Mar 30 '14 at 19:14
  • One thing to note though is that in $\mathbb{R}$ the eigen value of $a^2$ is $a^2$, and we need $|a^2|\lt 1$, but it's just not how you tend to think of things if we are not in $\mathbb{R^{n\times n}},n\gt 1 $. – Ellya Mar 30 '14 at 19:19
  • Does this mean that "since $([2I]^2 + I)^{-1} = \sum_{k=0}^\infty (-1)^k (2I)^{2k}$, and all the eigen values of $[2I]^2$ are NOT smaller than one, hence $([2I]^2 + I)^{-1}$ doesn't exist (and in particular it cannot be equal to $\frac{1}{5} I $)? – Calvin Lin Mar 30 '14 at 19:37
  • indeed, try multiplying $2I$ with itself infinite times and you will see that it diverges, by the way whats with the downvote?? – Ellya Mar 30 '14 at 20:35
  • @ellya: My +1 has nullified the downvote. –  May 05 '14 at 11:35
  • @LePressentiment thanks! didn't know why I was downvoted in the first place. – Ellya May 05 '14 at 11:36
  • @ellya: You're welcome! –  May 07 '14 at 11:36