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I am supposed to find the values of $a,b,c \in R$ for which the matrix can be diagonalized.

$\begin {bmatrix} 1 && 0 && c \\ 1 && a && b \\ 0 && 0 && 1 \end{bmatrix}$

I know it can be diagonalized if it has 3 linear independent eigenvectors, but that job seems too complicated in this case. I am not sure how to approach this problem, so any help is appreciated. Thanks

Ethan Bolker
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ivana14
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  • What do you know about diagonalizable matrices? What properties must a matrix have in order to be diagonalizable? In particular, what must be true about the eigenvalues/eigenvectors of a diagonalizable matrix? Based on that information, you should be able to solve this using the characteristic equation. – Vyas Jun 13 '17 at 15:47
  • @Vyas I added it to the question. I know that a matrix is diagonalizable iff it has $n $ linearly independent vectors. – ivana14 Jun 13 '17 at 15:49
  • You can start by computing eigenvalues. If you have 3 distinct eigenvalues, then you are already done. If you have <3 (which is the case here), you have to check on the dimension of the eigenspaces corresponding to the eigenvalues of multiplicity > 1. Does that help? – Vyas Jun 13 '17 at 15:54
  • Sure, I'll try that. Thanks – ivana14 Jun 13 '17 at 15:55
  • Implied in your question and your response to the comments is that you’re trying to diagonalize via an eigenbasis. There are other ways to diagonalize a matrix besides this. – amd Jun 13 '17 at 17:35
  • Yes, so I was wondering if there was an easier way to approach this problem.. – ivana14 Jun 13 '17 at 17:36
  • That depends on what sort of diagonalization is meant in the problem you’ve been given to solve. – amd Jun 13 '17 at 17:37

2 Answers2

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Hint:

Starting point:

$$\begin {vmatrix} 1-\lambda && 0 && c \\ 1 && a-\lambda && b \\ 0 && 0 && 1-\lambda \end{vmatrix} = (a-\lambda) \begin {vmatrix} 1-\lambda && c \\ 0 && 1-\lambda \end{vmatrix}=(a-\lambda)(1-\lambda)^2$$

You can consider two cases

Case 1. $a =1$ and

Case 2. $a \neq 1$.

Let $\lambda=1$ and see how many eigenvectors corresponds to it.

That is study the nullity of the matrix

$$\begin {bmatrix} 1-\lambda && 0 && c \\ 1 && a-\lambda && b \\ 0 && 0 && 1-\lambda \end{bmatrix}$$

Siong Thye Goh
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In order to diagonalize a matrix, you want to be able to utilize row operations such that you only end up with values along the diagonal. An easy case is $a = 1, c, b = 0$. Try plugging in and using row operations to simplify the matrix to see why and see if you can generalize this.

layabout
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