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I want to prove the following statement :

Let $B$ be a matrix, such that $B$ has the eigenvalue $0$ and no other eigenvalue. Then $B^2=0$.

In the context of the statement, $B$ is of size $2$. Is this hypothesis necessary for the statement to hold ?

How to prove the statement ?

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For sizes greater than 2x2, you can still say that the matrix must not be diagonalizable, or else its diagonalized form would be 0, so the matrix itself would have been 0. – Ben Crowell Mar 10 '12 at 17:53
up vote 8 down vote accepted

Let $B=\left[\matrix{ a&b\cr c&d\cr}\right]$

If $B$ has eigenvalue $0$ then it is not invertible, which means that det$B=ad-bc=0$.

The eigenvalues of a matrix are the roots of its characteristic polynomial

det$(B-\lambda I)=\lambda^2-(a+d)\lambda +(ad-bc)=\lambda(\lambda - (a+d))$

We know that the only eigenvalues are $0$, hence $a=-d$. Substituting this into the expression for the determinant, we get $bc = -a^2$. Now you can compute $\left[\matrix{ a&b\cr c&{-a}\cr}\right]^2=0$

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Thanks to all of you ! All the answers are excellent. – Henry Mar 10 '12 at 19:05
The computation for completeness: $\left[\matrix{a&b\cr c&{-a}}\right]\left[\matrix{a&b\cr c&{-a}}\right] = \left[\matrix{a^2+bc & ab-ba \cr ca-ac & cb+a^2}\right] = 0$. – ShreevatsaR Mar 11 '12 at 0:44

Yes the size condition on $B$ is necessary: take $B=\left[\matrix{ 0&1&1\cr 0&0&1\cr0&0&0}\right]$.

For $B$ a $(2\times 2)$ matrix, write down what the equation $\text{det}(B-\lambda I)=\lambda^2$ gives you. You'll deduce that both the trace and the determinant of $B$ are 0. So $B$ has the form $\left[ \matrix{a&b\cr c& -a}\right]$ where $a^2+bc=0$. Then $B^2= \left[ \matrix{a^2+bc&ab+b(-a )\cr ca+ (-a)c& cb+(-a)^2}\right] = \left[ \matrix{0&0\cr0& 0}\right]. $

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The characteristic polynomial $f$ of $B$ has degree 2 and since 0 is an eigenvalue of B it has 0 as a root, so $f$ splits into linear factors: $f(X) = X(X-a)$ for some $a$. Since 0 is the only eigenvalue we must have $a=0$, hence $f(X) = X^2$. By the Cayley-Hamilton theorem, $0 = f(B) = B^2$.

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Nice. I TAed a first year lin. alg. course last term, so I forgot about the Caley-Hamilton theorem :P – you Mar 10 '12 at 16:58

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