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Question:

Give an orthonormal basis for null(T), for $ T \in \mathbb{L(\mathbb{C^4)}}$ is the map with canonical matrix: $$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ \end{bmatrix} $$

My Steps:

First I put the matrix in RREF to find the dimension of the null space: $$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} $$ Resulting in a null space dimension 3. With this I solved the matrix for the null space getting $$ x_1 = -a -b -c $$ $$ x_2 = a $$ $$ x_3 = b $$ $$ x_4 = c $$ Solving for the basis of the null space, I got $$ a \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix} + b \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \\ \end{bmatrix} + c \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \\ \end{bmatrix} $$ Was this the proper way to solve for the basis of the null space given this matrix and if so, since a, b, and c are arbitrary scalars, could I apply the Gram-Schmidt procedure on the three vectors to find the orthonormal basis?

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    $\begingroup$ Yes, go ahead with Gram-Schmidt. $\endgroup$ – Friedrich Philipp Nov 11 '17 at 23:41
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I agree with the comment that Gram-Schmit process is the right way to go.

Alternative approach to a quick solution exists if you are familiar with Hadamard matrix.

$$H_4H_4^T = 4I$$

where $$H_4 = \begin{bmatrix} 1 & 1 & 1 & 1 \\ -1 & 1 & -1 & 1 \\ -1 & -1 & 1 & 1 \\ 1 & -1 & -1 & 1 \\ \end{bmatrix}$$

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