matrix diagonalization with unknowns

Question

For the matrix: $$A=\begin{pmatrix} 0 & 0 & 0 & 0\\ x & 0 & 0 & 0\\ 0 & y & 0 & 0\\ 0 & 0 & z & 0\\ \end{pmatrix}$$

I need to find values fro x,y,z that allow the matrix to be diagonalized.

Do I need to swap the rows, moving R1 to the bottom to make the diagonal non-zero. If this is the case what do I do from there

Answer 1

Well we know that the the eigenvalues are zero, simply because the matrix is upper triangular and the diagonals are all zero. Thus we get our eigenvalues by taking the nullspace of $A$. The matrix is diagonalizable in this case if and only if there are at least four linearly independent eigenvectors, that is $\dim(\ker(A))\ge 4$. So at this point you should look at what values of $x,y,z$ cause $A$ to have a kernel of dimension less than four, think about rank.

Answer 2

We get that the characteristic polynomial of matrix A is: $C_A(x)= x^4$ so the eigenvalues are 0,0,0,0. Now we know that A is diagonalizable iff $dimV_A(0)=4$. Now remembering that $dimV_A(0)=dim \mathbb R^{4×1}-rank(A-0I_4)$ we conclude that A is diagonalizable iff: $rankA=0$

Answer 3

Your characteristic polynomial is $P=X^4$, so the only possible eighen values is $0$. It means that if your diagonalize your matrix, its diagonalized version is $0$.

So you will have $A=P^{-1}DP=P^{-1}\cdot 0 \cdot P=0$. That imply it is possible only if $A=0$ i.e. $x=y=z=0$.

You can also prove that your matrix is nilpotent (Cayley-Hamilton's theorem, or simply by seeing that $A^4=0$) and that any nilpotent matrix is diagonalizable if it is $0$.

Demo : A nilpotent ($A^n=0$), if it is diagonalizable we know the existence of $P$ with P not null so that $A=P^{-1}DP$ and $A^n=P^{-1}D^nP=0$ so $D^n=0$. For a diagonal matrix that implies that it is null.

Answer 4

The characteristic polynomial of

$$A=\begin{pmatrix} 0 & 0 & 0 & 0\\ x & 0 & 0 & 0\\ 0 & y & 0 & 0\\ 0 & 0 & z & 0\\ \end{pmatrix}$$ is $\lambda^4$. Thus $0$ is an eigenvalue with multiplicity 4. The eigenspace associated to eigenvalue $0$ should thus be 4-dimensional.

Let us remark that this eigenspace is plainly the kernel (null space) of $A$.

But the kernel is at most 3-dimensional in the usual cases (i.e. for a non-zero matrix) due to the rank-nullity theorem (https://en.wikipedia.org/wiki/Rank–nullity_theorem). If for example $x \neq 0$, the first column of $A$ is a non nul vector, element of the range of $A$, thus rank$(A) \geq 1.$

Conclusion: $A$ is diagonalizable in the unique case $x=y=z=0.$

Answer 5

The matrix satisfies $A^4=0$ independently of $x,y,z$, so its minimal polynomial always divides $X^4$. Only if the minimal polynomial has simple roots is the matrix diagonalisable; given the divisibility this can only be if the minimal polynomial is $X$. Thus $A$ is diagonalisable if and only if $A=0$.