# Give an orthonormal basis for null(T), for $T \in \mathbb{L(\mathbb{C^4)}}$

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?

• Yes, go ahead with Gram-Schmidt. – Friedrich Philipp Nov 11 '17 at 23:41

$$H_4H_4^T = 4I$$
where $$H_4 = \begin{bmatrix} 1 & 1 & 1 & 1 \\ -1 & 1 & -1 & 1 \\ -1 & -1 & 1 & 1 \\ 1 & -1 & -1 & 1 \\ \end{bmatrix}$$