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1) If $AB+BA=0$, then $A^2B^3=B^3A^2$

2) If $A$ and $B$ are non-singular, then $AB$ is non-singular

3) If $A^3=0$, then $A-I$ is non-singular

$A$ and $B$ are $n \times n$ matrices and $I$ is identity matrix.

And for 3), If square of matrix is $0$, then its matrix has to be zero element?

Edward Jiang
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aaa
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  • Number 2 is easy if you think about matricies as linear transformations. – Matthew Levy Oct 13 '14 at 03:39
  • pjhuxford - I tried everything and wanted to see if I did right. Don't regard me as someone who posts up problems they never tried – aaa Oct 13 '14 at 03:48

1 Answers1

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For 1, $AB+BA=0\implies AB=-BA$. Using this, we can try and turn the LHS $A^2 B^3$ into $B^3A^2$ by succcessively replacing $AB$ with $-BA$.

\begin{align*} A^2B^3 &= A(AB)B^2 \\ &= A(-BA)B^2 \\ &= -(AB)(AB)B \\ &= -(-BA)(-BA)B \\ &= -B(AB)(AB) \\ &= -B(-BA)(-BA) \\ &= -B^2(AB)A \\ &= -B^2(-BA)A \\ &= B^3A^2\:. \end{align*}

For 2, a matrix is nonsingular iff its determinant is nonzero. If we know that $\det A,\det B\neq0$, this implies that

$$\det(AB)=(\det A)(\det B)\neq0.$$

For 3, we take inspiration from the factorisation of $x^3-1=(x-1)(x^2+x+1)$. Since $A^3=0$, this means that $I-A^3=I$. It can be verified by expansion that

$$(A-I)(-A^2-A-I)=(-A^2-A-I)(A-I)=I-A^3=I$$

which implies that $A-I$ is invertible, and has inverse $-A^2-A-I$.

Peter Huxford
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