Why is this matrix not diagonalizable 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!
$\begin{bmatrix}
1 & 1 \\
0 & 1 
\end{bmatrix}$
 A: 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.
A: 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, $e_1$.
