Find the value of $k$ for which matrix is diagonalizable

Question

Consider the matrix

$$A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & k \\ 0 & 0 & 2 \\ \end{bmatrix}$$ where $k$ is a real number.

The characteristic polynomial of $A$ is $(λ −1)(λ−2)^2$. Find the value of $k$ for which the matrix $A$ is diagonalizable and write down a matrix $P$ which diagonalizes $A$ for this value of $k$.

I have attempted the problem by trying to find eigenspaces and hence the basis, to get the matrix $P$. Tried row reducing and other things, but can't seem to get anywhere. I feel like there is a theorem which relates the characteristic polynomial and diagonalizable matrix, but I'm not sure. Any help would be greatly appreciated!

Answer 1

Right, we can read off the characteristic polynomial from the diagonal: $$p(\lambda) = (\lambda-1)(\lambda-2)^2.$$Call the matrix $A$. If the matrix is to be diagonalizable, we must obtain a basis of eigenvectors. We need three vectors. In other words, since $$\dim(\ker(A-{\rm Id})) \geq 1, \quad \dim(\ker(A-2\,{\rm Id})) \geq 1,$$ and $1$ is a simple root of $p(\lambda)$, we have $\dim(\ker(A-{\rm Id}))=1$. So $\dim(\ker(A-2\,{\rm Id}))$ had better be $2$. $$A-2\,{\rm Id } = \begin{bmatrix} -1 & 0 & 1 \\ 0 & 0 & k \\ 0 & 0 & 0 \\ \end{bmatrix}$$ From this it is clear that we get what we want for $k=0$.

Answer 2

Matrix is Diagonalizable if and only if minimal polynomial is of the form

$$p(x)=(x-c_1)(x-c_2)...(x-c_k)$$

where $c_1,c_2,...,c_k$ are distinct eigen values of matrix.

Here in your case matrix will be diagonalizable if and only if $$(A-\,{\rm I })(A-2\,{\rm I })=0$$ $$(A-\,{\rm I })(A-2\,{\rm I }) = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & k \\ 0 & 0 & 1 \\ \end{bmatrix}\begin{bmatrix} -1 & 0 & 1 \\ 0 & 0 & k \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$$=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & k \\ 0 & 0 & 0 \\ \end{bmatrix}$$. Hence for $k=0$ matrix is diagonalizable.