Given the matrix $A= \begin {pmatrix} 1 & 1 &1 \\ -1 & 1 & 0 \\ 0 & 2 &1 \end{pmatrix}$.
(i) Determine the orthogonal projection $p:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ on $Im(A)$
(ii) Calculate an orthonormal basis of $(\ker(A))^{\perp}$
(iii) Determine the pseudoinverse $A^+$ of $A$
I was wondering about the sequence of the subtasks. Normaly, i would do (iii) first and then (i) & (ii) using that $AA^+:\mathbb{R}^m\rightarrow Im (A)$ and $A^+A:\mathbb{R}^n\rightarrow(\ker(A))^{\perp}$. So my question is: Is there a way to to (i) and (ii) without determine the pseudoinverse?
Best Answer
There sure is.
For (i), row reduce (or find a suitable linear combination) to find that $Im(A) = span(v_1,v_2)$, where $v_1$ and $v_2$ are the columns of $A$. From there, we want to find the orthogonal projection onto the space spanned by these two vectors.
In order to do that, we would find an orthonormal basis (using Gram-Schmidt) $u_1,u_2$ of the span. In this case, we find $$ u_1 = \frac 1{\sqrt 2}\pmatrix{1\\-1\\0}, \quad u_2 = \frac 1{\sqrt 6}\pmatrix{1\\1\\2} $$ We then have have $$ P = u_1 u_1^T + u_2 u_2 ^T = \frac 12\pmatrix { 1&-1&0\\ -1&1&0\\ 0&0&0 } + \frac 16 \pmatrix { 1&1&2\\ 1&1&2\\ 2&2&4 } $$ For (ii): an orthonormal basis for $\ker(A)^\perp$ is equivalently an orthonormal basis for $Im(A^T)$. So, find a basis for $Im(A^T)$, then apply the Gram Schmidt process to make an orthonormal basis for the same span.