2

Let $A$ be an $n\times n$ anti-diagonal matrix: $a_{i,j}=0$ unless $i+j=n+1$.

A) When is $A$ diagonalizable (what are the conditions on the $a_{i,n+1−i}$)?

B) Find the eigenvalues and eigenvectors of $A$ for $n=7$ and $a_{i,n+1−i}=i^2$.

Not sure how to get started on part (A) - I've played around with $\det[A-\lambda I]$ for $A$, an $n\times n$ anti-diagonal and didn't come up with any specific conditions for diagonalizability.

I know we need to produce a basis of eigenvectors of $A$ -- showing that $A$ is similar to a diagonal matrix $D$.

For part (B), all the anti-diagonal entries take the value $i^2 = -1$. I'm currently trying to compute this by brute force - and got a bunch of -$\lambda$'s on the main diagonal, $-1-\lambda$ in the middle entry of the matrix, and then $\det[A-\lambda I]$ is something like $p(\lambda)$ = $-\lambda^7 - \lambda^{6}$. Solving for the eigenvalues (and then computing the corresponding eigenvectors) doesn't really follow at this point, so I'm guessing there is a trick that I need to use.

Thanks in advance,

Xoque55
  • 4,499
User001
  • 1

1 Answers1

1

if you make $A$ symmetric, then $A$ would certainly be diagonalizable. that is you make $a_{1n} = a_{n1}, a_{2n-1} = a_{n-12}, \cdots.$ it is enough for matrix $A$ to be normal, i.e., $AA^* = A^*A,$ where $A^*$ is the complex conjugate of $A^T,$ for it to be diagonalizable. if you look at a $3 \times 3$ matrix $A = diag(a, b, c)\ $, you see that $AA^* = diag(|a|^2, |b|^2, |c|^2), A^TA = diag(|c|^2, |b|^2, |a|^2)$ now you only need that $|a| = \pm |c|$ for $A$ to be diagonalizable over $C$ which is less stringent than $a = c$

see if you can verify that $|a_{1n}| = \pm |a_{n1}|, |a_{2n-1}| = \pm |a_{n-12}|, \cdots$ is necessary and sufficient for $A$ to be diagonlizable.

edit:

hint for part (b). i will use $e_1 = (1,0,0,0,0,0,0)^T,\cdots$ for the standard basis vectors in $R^7.$

here we have $n = 7$ and $Ae_1 = 49e_7, A e_2 = 36e_6,\cdots, Ae_6 = 4e_2, Ae_7 = e_1$ observe the pairing $(Ae_1, Ae_7), (Ae_2, Ae_6), \cdots$ and a singleton $Ae_4 = 16 e_4$ last one already tells you that $4$ is an eigenvalue and a corresponding eigenvectors is $e_4.$ the rest make up three $2 \times 2$ blocks.

can you find the eigenvalues and the corresponding eigenvectors for each of the three pairs?

abel
  • 29,612
  • 2
    No, it isn't as $$\begin{pmatrix}0&1\!!-1&0\end{pmatrix}$$ isn't diagonalizable (over the reals) – Timbuc Jan 04 '15 at 03:46
  • @Timbuc, $AA^T = A^TA$ does not imply $A$ is diagonalizable over $R.$ so i should change the condition to $A^* A = AA^*$ and make it diagonalizable over $C?$ – abel Jan 04 '15 at 03:57
  • I know that. Anyway, it may depend on what the OP wanted, as he didn't specified field. – Timbuc Jan 04 '15 at 04:15
  • Awesome, @abel :) I have to get better at un-disguising these problems - turns out this was just an exercise in applying the properties of normal matrices over C. I'll work on this a bit on paper now; I was reviewing some block matrices stuff just before. If you have some time, can you give me a hint on part (b)? I'm sure there must be a trick, and that I am not really being asked to compute a 7x7 determinant nor solve a 7-degree polynomial in $\lambda$. (The campus library is closing in 10 minutes, so I might not be on this site actively tonight...though I can check my phone) thanks!! :) – User001 Jan 04 '15 at 04:49
  • Hi @Abel, just revisiting part b of this question and was wondering ... how'd you get those scalars 49...36...and 4, etc? We don't have a specific matrix A of coefficients to apply to the standard basis vectors that you chose to work with - we only know that it is anti-diagonal. Thanks ... – User001 Jan 06 '15 at 02:21
  • Actually $i^2$ on the antidiagonal and zero elsewhere - but I still don't see where the scalars are coming from ... – User001 Jan 06 '15 at 02:23
  • 1
    @LebronJames, the index stands for the number of the row. on the first row you have $1$ and on the second one you have $4$ and so on. read the problem carefully. write out the elements on a piece of paper instead reading a screen. – abel Jan 06 '15 at 02:31