Use the Gram-Schmidt process to find the orthonormal basis for the row space of the matrix $A$.
The matrix $A$ is as follows:
\begin{bmatrix}1&1&0&0\\-1&3&0&1\\-3&1&-2&1\end{bmatrix}
I have worked out the solution as follows:
Step 1: Finding the row space.
The rows of A are
$r_1 = \begin{bmatrix}1&1&0&0\end{bmatrix}$
$r_2 = \begin{bmatrix}-1&3&0&1\end{bmatrix}$
$r_3 = \begin{bmatrix}-3&1&-2&1\end{bmatrix}$
Hence, the row space of $A$
= span {$r_1, r_2, r_3$}.
= $ ${ $ a\begin{bmatrix}1&1&0&0\end{bmatrix} + b\begin{bmatrix}-1&3&0&1\end{bmatrix} + c\begin{bmatrix}-3&1&-2&1\end{bmatrix} $}
= { $ \begin{bmatrix}a-b-3c&a+3b+c&-2c&b+c\end{bmatrix}$ }
Step 2: Finding the basis of the row space. The rref of matrix $A$ is:
\begin{bmatrix}1&0&0&-1/4\\0&1&0&1/4\\0&0&1&0\end{bmatrix}
so the basis of $A$ is:
{ $\begin{bmatrix}1&0&0&-1/4\end{bmatrix}, \begin{bmatrix}0&1&0&1/4\end{bmatrix}, \begin{bmatrix}0&0&1&0\end{bmatrix} $ }
Step 3: Gram-Schmidt Process
$v_1 = \begin{bmatrix}1&0&0&-1/4\end{bmatrix}$
$v_2 = \begin{bmatrix}1/17&1&0&9/34\end{bmatrix}$
$v_3 = \begin{bmatrix}0&0&1&0\end{bmatrix}$
Then { $\begin{bmatrix}1&0&0&-1/4\end{bmatrix}, \begin{bmatrix}1/17&1&0&9/34\end{bmatrix}, \begin{bmatrix}0&0&1&0\end{bmatrix}$ } is an orthogonal basis for the row space of $A$.
The orthonormal basis is then as follows:
{ $ \sqrt{16/17} \begin{bmatrix}1&0&0&-1/4\end{bmatrix}, \sqrt{68/73} \begin{bmatrix}1/17&1&0&9/34\end{bmatrix}, \begin{bmatrix}0&0&1&0\end{bmatrix}$ }
Is this solution correct?
Best Answer
Let me get something out of the way first: there’s no such thing as the orthonormal basis of a nontrivial real vector space. Indeed, if the space is at least two-dimensional, there is an infinite number of such bases. By the same token, there’s no such thing as the basis of the row space of a matrix. What you’ve found by row-reduction is only one possible basis. For instance, from that very same RREF matrix, I could instead read the basis $\{[4,0,0,-1],[0,4,0,1],[0,0,1,0]\}$. Also, since the matrix obviously has full rank, the rows of the original matrix form a basis for its row space, too. What this means in practice is that there is not a single correct answer to this exercise. If you’ve come up with a set of pairwise orthogonal unit vectors that span the space, that’s a correct answer.
Checking orthogonality, though, there’s a problem: the first two vectors in the basis that you came up with are not orthogonal. $\left[1,0,0,-\frac14\right]\cdot\left[\frac1{17},1,0,\frac9{34}\right] = -\frac1{136}\ne 0.$ You should double-check your work for that step of the G-S process.
You could’ve saved yourself a lot of work, though, by simply applying the Gram-Schmidt process directly to the rows of $A$. If at any point it produces a zero vector, that just means that the input to that step is a linear combination of the already-processed vectors, so you can simply discard it and continue with the next one. For this matrix, the values that you end up working with turn out to be much “nicer” than the ones that you get with the basis derived from the RREF of $A$. Proceeding without normalization, as you did, we obtain the orthogonal basis $\{[1,1,0,0],[-2,2,0,1],[0,0,-2,0]\}$, which also normalizes to “nice-looking” numbers.