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Let $A$ be a square matrix $A^2=0$ and $A\neq0$ and show that it is not diagonalizable.

I decided to use the sample matrix of $$A = \begin{bmatrix}0 & 1\\0 & 0 \end{bmatrix}$$ which satisfies the conditions above.

So my question is: how would I prove this is not diagonalizable. The matrix leads to a eigenvalue of $\lambda=0$ with an algebraic multiplicity of $2$.

I know that if the algebraic multiplicity and geometric multiplicity are equal, then it is diagonalizable.

But I am kind of stuck from here since when I use $\det(A-0I)=0$, it just leads to get $x_2=0$ but then also that $x_2=t$ so I don't really know what to do. Any help would be appreciated.

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  • $\begingroup$ If the $2\times 2$ matrix with only the eigenvalue zero had two linearly independent eigenvectors (geometric multiplicity two), that matrix would be the zero matrix. $\endgroup$
    – hardmath
    Jan 12, 2015 at 22:55

5 Answers 5

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It isn’t enough to prove that your particular sample matrix isn’t diagonalizable: you must show that every non-zero square matrix $A$ such that $A^2=0$ is non-diagonalizable.

HINT: Suppose that $A^2=0$ and $A$ is diagonalizable. Then there are an invertible matrix $P$ and a diagonal matrix $D$ such that $D=P^{-1}AP$.

  • What is $D^2$?
  • What does this tell you about $A$?
  • How does this prove the desired result?
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The only eigenvalue is 0. If it's diagonalizable, it is similar to the zero matrix, so it's equal to the zero matrix. But it's not.

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first the eigenvalues of $N = \pmatrix{0&1\\0&0}$ are $0,0$ if $N$ were digonalizable, then the diagonal matrix must be the zero matrix. so $UDU^-1$ will be the zero matrix too. therefore it cannot be $N$ and $N$ is not diagonalizable.

the same argument holds for any matrix $A \neq 0$ with $A^2 = 0$ reason is $A^2 = 0$ implies $0$ is the only eigenvalue.

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If $A^k=0$ (in your case with $k=2$), then any eigenvalue of $A$ satisfies $\lambda^k=0$ (check what happens to an eigenvector) and therefore $\lambda=0$. Then if $A$ is diagonalisable, the only possibility is that $A=0$.

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You have $det(A^2)=0$ then $det(A)=0$. This implies A has a zero eigenvalues which gives you that A is not diagonalizable

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