I'm studying for an exam and I'm not sure how to do this. I understand what the definitions mean (for the most part) but I'm not sure how to apply it to the problem.
Let
A = \begin{pmatrix}1/2&-1/2\\1/2&-1/2\\1/2&1/2\\1/2&1/2 \end{pmatrix}
a) Find an orthonormal basis for Null( A$^T$ )
and
b) Determine the projection matrix Q that projects vectors in $\mathbb{R}$$^4$ onto Null(A$^T$).
My thoughts:
The matrix's column vectors are definitely orthonormal, so I want to find a basis such that for any x, Ax = 0.
For b) I want to use the projection formula and find some vector b within the above basis?
Help / an explanation of steps would be appreciated, thanks.
Best Answer
Solution To Part (a)
I’ll go through the general method, although in this case you can almost eyeball the solution.
One way to find a basis for the kernel (a.k.a. nullspace) of a matrix is to use row-reduction. Row-reducing $A^T$ goes pretty quickly since there are only two rows, producing $$ R=\pmatrix{1&1&0&0\\0&0&1&1} $$ Find the columns that don’t have leading entries, in this case, the second and last. Basis vectors for the kernel will have a one on one of the rows that corresponds to these positions and zeroes in the others, so we’ll have $\langle a,1,b,0\rangle$ and $\langle c,0,d,1\rangle$ as our basis vectors. Formally, you’d now solve for the missing components, but in practice you can just read them from the row-reduced matrix:$$ R\pmatrix{a\\1\\b\\0}=\pmatrix{a+1\\b}=0, $$ so $a=-1$ and $b=0$. But these values are just the second column of $R$ negated. Similarly, the second basis vector will be $\langle0,0,-1,1\rangle$. These vectors are obviously orthogonal, so all you need to do is normalize them, yielding $$ \pmatrix{-\frac1{\sqrt2}\\\frac1{\sqrt2}\\0\\0}\text{ and }\pmatrix{0\\0\\-\frac1{\sqrt2}\\\frac1{\sqrt2}}. $$
Hint For Part (b)
Use the fact that projection onto $\operatorname{ker}A^T$ can be broken down into projections onto its basis vectors.