If v ∈ R^n is a nonzero vector, and I ∈ R^n×n is an identify matrix. Prove that M = I − 2(v(v^T)/||v||^2 is symmetric and satisfies M^−1 = M

Question

I thought about showing M = M^T, so M is symmetric. But I don't know how to compute 2(v(v^T)/||v||^2 to find M and M^T. Any idea?

Answer 1

To prove it is symmetric just compute the transpose using the well known properties $(A+B)^T=A^T+B^T$ and $(\lambda A)^T=\lambda A^T$. It should not be a problem.

As for the second part, just multiply $M$ by itself and prove you get the identity matrix. Remember that $v^Tv=\langle v,v\rangle=||v||^2$, you will have to use that.

Answer 2

I would first show that symmetric matrices are closed under addition and scalar multiplication. Then I would show that $I$ is symmetric and $v v^T$ is symmetric since they are the non-scalar parts of the equation. To see that $M=M^{-1}$ show that $M^2=I$ and you're done.

Answer 3

1

Pf that $M$ is Symmetric i.e. $M^T = M$

$$M=I-2\frac{vv^T}{|v|^2}$$

$$\implies M^T=(I-2\frac{vv^T}{|v|^2})^T$$

$$=(I + -2\frac{vv^T}{|v|^2})^T$$

$$=(I)^T + (-2\frac{vv^T}{|v|^2})^T$$

$$=I + (-2\frac{vv^T}{|v|^2})^T$$

$$=I + -2\frac{1}{|v|^2} (vv^T)^T$$

$$=I + -2\frac{1}{|v|^2} (v^T)^T (v)^T$$

$$=I + -2\frac{1}{|v|^2} v(v)^T$$

$$=I-2\frac{vv^T}{|v|^2} = M$$

Therefore, $M^T=M$. QED

2

Pf that $M^2=I$:

$$M=I-2\frac{vv^T}{|v|^2}$$

$$\implies M^2=(I-2\frac{vv^T}{|v|^2})(I-2\frac{vv^T}{|v|^2})$$

$$=I^2 + 2I(-2\frac{vv^T}{|v|^2}) + (2\frac{vv^T}{|v|^2})(2\frac{vv^T}{|v|^2})$$

$$=I + 2I(-2\frac{vv^T}{|v|^2}) + (2\frac{vv^T}{|v|^2})(2\frac{vv^T}{|v|^2})$$

$$=I + 2I(-2\frac{vv^T}{|v|^2}) + 4\frac{vv^T}{|v|^2}\frac{vv^T}{|v|^2})$$

$$=I + 2(-2\frac{vv^T}{|v|^2}) + 4\frac{vv^T}{|v|^2}\frac{vv^T}{|v|^2})$$

$$=I -4 \frac{vv^T}{|v|^2}) + 4\frac{vv^T}{|v|^2}\frac{vv^T}{|v|^2})$$

$$=I -4 \frac{vv^T}{|v|^2}) + 4\frac{vv^Tvv^T}{|v|^2|v|^2}$$

$$=I -4 \frac{vv^T}{|v|^2}) + 4\frac{v(v^Tv)v^T}{|v|^2|v|^2}$$

$$=I -4 \frac{vv^T}{|v|^2}) + 4\frac{v|v|^2v^T}{|v|^2|v|^2}$$

$$=I -4 \frac{vv^T}{|v|^2}) + 4\frac{vv^T}{|v|^2}$$

$$=I$$

QED

3

Pf that $M^{-1}$ exists:

Recall that a matrix $A$ has an inverse if and only if its determinant is nonzero.

$$M^2=I$$

$$\implies \det(M^2) = \det(I)$$

$$\implies \det(M)^2 = \det(I)$$

$$\implies \det(M)^2 = 1$$

$$\implies \det(M) = \pm 1$$

$$\implies \det(M) \ne 0$$

QED

Edit: I have no idea what was going on but of course $M^{-1}$ exists a fortiori if $M^2=I$ because $M^2=I$ is equivalent to saying '$M^{-1}$ exists and $M^{-1}=M$.'

P.S. Good luck with probability or statistics next!