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Im studying for an exam in linear algebra, and doing some false/true statements, and can't figure out why this matrix is not diagonalizable. I am supposed to see that without any calculations.

Would anyone mind explain? Thanks!

$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$

uoiu
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    There's this: https://math.stackexchange.com/questions/1558591/if-a-matrix-is-triangular-is-there-a-quicker-way-to-tell-if-it-is-can-be-diagon both diagonal entries in your matrix are equal and thus it would only be diagnosable if it was already a diagonal matrix which it is not. I will admit however that I was unaware of this general result. – TC159 Dec 19 '22 at 14:00
  • Thanks! That helped me a lot!! – uoiu Dec 19 '22 at 14:03
  • What is the characteristic polynomial of this matrix, and what are the geometric and algebraic multiplicities of the eigenvalue? – Irving Rabin Dec 19 '22 at 14:03
  • This particular matrix is a Jordan block (or Jordan cell) of size bigger than $1\times1$, which even tells you what its Jordan chain will look like. – nasekatnasushi Dec 19 '22 at 14:05
  • They were supposed to do this without calculations @IrvingRabin, in fact I had a written answer which I had to delete after rereading the question. – TC159 Dec 19 '22 at 14:05
  • @nasekatnasushi I doubt OP is expected to know that beforehand. – TC159 Dec 19 '22 at 14:07
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    If you studied Jordan's canonical form, it should be clear that this is a Jordan block, and therefore not diagonizable. – PierreCarre Dec 19 '22 at 14:09
  • @nasekatnasushi Unfortunately, I don't recognise Jordan block. – uoiu Dec 19 '22 at 14:09
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    It's a true or false question @TC159. I generally have some scratch work to situate myself on a true or false. Having the result you linked in the back pocket is nice but a bit specific imo. At least they would know where to go if they didn't see it immediately. – Irving Rabin Dec 19 '22 at 14:10
  • Similarity preserves eigenvalues. If your matrix is diagonalisable (i.e., if it is similar to a diagonal matrix), it will be similar to a diagonal matrix whose eigenvalues are all equal to $1$, which is the identity matrix $I$. But this is impossible, because $I$ is only similar to itself. – user1551 Dec 19 '22 at 14:47

2 Answers2

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Even if this is a True/False question, it does not mean that you cannot do any calculations...

Since $A$ is upper triangular, you immediately know that $\lambda = 1$ is the only eigenvalue (with algebraic multiplicity 2). Now, the matrix would be diagonizable if the corresponding eigenspace had dimension 2. However, the general solution of $(A- I)u = 0$ is just $(u_1,0)$, $u_1 \in \mathbb{R}$, which means that the dimension of the eigenspace is one and the matrix is not diagonizable.

PierreCarre
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Denote by $e_1, e_2$ the basis for the $2$-dimensional vector space you're working on.

Calling your matrix $A$, the first thing you can observe is that $Ae_1 = e_1$, so you have already found one eigenvector + eigenvalue.

Now, two scenarios:

  • either you have already studied triangular matrices, and you know that the second eigenvalue of $A$, if it has one, also has to be $1$
  • or you haven't, in which case we will just assume $A$ does have a second eigenvalue $\lambda$, associated to some vector $v$ (ie $Av = \lambda v$)

The second column of your matrix $A$ tells you that $Ae_2 = e_1 + e_2$.

Now, writing $v = ae_1 + be_2$, for arbitrary scalars $a,b$, you have:

$Av = ae_1 + be_1 + be_2 = (a+b)e_1 + be_2$, by writing out the calculations.

So, $Av = \lambda v \iff (a+b)e_1 + be_2 = \lambda(ae_1 + be_2)$

In other words, you need to satisfy the two equations $a + b = \lambda a$ and $b = \lambda b$.

Clearly, the only possible solution to this system is for $\lambda = 1, b = 0$. So then, the only possible choice of eigenvector is $ae_1$ for $a$ a scalar.

So the only eigenvector of your matrix was the obvious one, $e_1$.

Azur
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