Vtyjkyuk

If the $ntimes n$ square matrix $A$ satisfy $operatornamerank(A+3E)+operatornamerank(A-3E)=n$, $E$ is the identity matrix , prove A is similar to a diagonal matrix $operatornamediagx_1,x_2,cdots,x_n $ where $x_i=3$ or $-3$ for $i=1,2,cdots,n$.

I have been told to get this by discussing the geometric multiplicity of $lambda=3$ and $lambda=-3$ . But I am not familiar with the properties of geometric multiplicity. Any more elementary method to get the point?

From $operatornamerank(A+3E)+operatornamerank(A-3E)=n$ and the nullity-rank theorem we get

$n= dim (ker(A+3E))+dim (ker(A-3E))$. Hence, if $ mathbb K= mathbb R$ or$= mathbb C$:

$mathbb K^n= ker(A+3E) oplus ker(A-3E).$

Can you proceed ?

Hint: If $A$ is similar to a diagonal matrix $diag(3,dots 3, -3, dots, -3)$, then there are vectors $b_1,dots, b_k,c_1,dots,c_l$ which form a basis such that $A b_i = 3b_i$ and $Ac_i = -3c_i$. Those vectors are eigenvectors. So you have to assert that there are enough of those vectors.

Since $DeclareMathOperatorKerKerDeclareMathOperatordimdim dim(Ker(A+3E))=n-operatornamerank(A+3E)$ and $dim(Ker(A-3E))=$ $n-operatornamerank(A-3E)$, we have
$$dim (Ker(A-3E))+dim(Ker(A+3E))=n.$$
The subspaces $Ker(A-3E)$ and $Ker(A+3E)$ intersect only at $0$ and their dimensions add up to $n$. So we can choose a basis $v_1,ldots, v_n$ for $mathbbR^n$ such that $v_1,ldots, v_k$ is a basis for $Ker(A-3E)$ and $v_k+1,ldots, v_n$ is a basis for $Ker(A+3E)$. Then the matrix of $A$ in this basis is a diagonal matrix with diagonal entries equal to $3$ or $-3$.