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Matrix: \begin{bmatrix} 4& -4 &4 \\ -3& 3 &5 \\ 1& -1&9 \end{bmatrix}

I want to check whether this matrix is diagonalizable and thus I want to calculate the eigenvalues. I know that this is possible with cofactor expansion and I have tried it and came to the right answer, but is there any FASTER way of getting eigenvalues of such matrix?

cmk
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NaN
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  • If you're allowed to use software like Wolfram|Alpha, that's going to be pretty fast: EigenSystem[{{4,-4,4},{-3,3,5},{1,-1,9}}]. – Adrian Keister Jul 03 '19 at 13:09
  • yes I know that I can do it with Wolfram but I don't think that is allowed in an exam setting – NaN Jul 03 '19 at 13:10
  • @AnuragA good observation didn't think of that, thanks. – NaN Jul 03 '19 at 13:16
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    Well, if you're in an exam setting, that would be good context to include in the original question, because it limits the kinds of answers you're willing to accept. Otherwise, you're wasting people's time. Here at M.SE, we love context! – Adrian Keister Jul 03 '19 at 13:18

1 Answers1

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Since the first two columns are multiples of each other, so $\lambda=0$ is definitely an eigenvalue. Also $\text{trace}=16$, so sum of the other two eigenvalues is $16$.

Also observe that (by adding the second and third column) $$A\begin{bmatrix}0\\1\\1\end{bmatrix}=8\begin{bmatrix}0 \\ 1\\ 1\end{bmatrix}.$$

Thus $\lambda=8$ is a repeated eigenvalue (using the trace idea).

Anurag A
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  • Is it always the case that using the trace idea the sum/2 is an eigenvalue? – NaN Jul 03 '19 at 13:19
  • @NaN Trace is equal to the sum of all the eigenvalues. Since one of them happens to be $0$, therefore the the other two add up to the trace (in this case). – Anurag A Jul 03 '19 at 13:20
  • but there is no way of knowing whether the other 2 eigenvalues are actually identical right? because there is no unique solution for sum = 16 – NaN Jul 03 '19 at 13:21
  • @NaN Yes they needn't be identical. But in this problem, by the argument I showed above, one of them turned out to be $8$ so with trace being $16$, the other will also be $8$. – Anurag A Jul 03 '19 at 13:22
  • nevermind I see what you did, but how did you know adding those columns up would result in knowing one of the 2 unknown eigenvalues ? – NaN Jul 03 '19 at 13:25
  • @NaN It just so happens that this matrix had that property. Since multiplying a matrix be a vector essentially means taking a linear combinations of columns of the matrix so I tried that approach and it worked. – Anurag A Jul 03 '19 at 13:28
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    Ah ok thanks for the effort – NaN Jul 03 '19 at 13:33