# Let $q_1,q_2,q_3,q_4$ be orthonormal vectors in $\mathbb{R}^4$,$z_1,z_2,…,z_6$ orthonormal vectors in $\mathbb{R}^6$ and $A=z_1q_1^T + z_2q_2^T$.

Let $q_1,q_2,q_3,q_4$ be orthonormal vectors in $\mathbb{R}^4$,$z_1,z_2,...,z_6$ orthonormal vectors in $\mathbb{R}^6$ and $A=z_1q_1^T + z_2q_2^T$.

a) Find the base and dimension of fundamental subspaces of matrix A

b) Using the results given by a) find the general solution for $Ax - z_1$.

Any hint would be appreciated.

Note that $A \in M_{6 \times 4}(\mathbb{R})$ and
$$Ax = z_1 q_1^T x + z_2 q_2^T x = z_1 (q_1^T x) + z_2 (q_2^T x) = \left< q_1, x \right> z_1 + \left< q_2, x \right> z_2$$
where $\left< \cdot, \cdot \right>$ is the standard inner product on $\mathbb{R}^4$. Hence, every vector in the image of (the linear map defined by) $A$ is a linear combination of $z_1$ and $z_2$. By taking $x = q_1,q_2$ and using the orthogonality of $q_i$ we see that $z_1,z_2$ belong to the image of $A$. Hence $\operatorname{rank} A = 2$ and $\dim \ker A = 4 - 2 = 2$.
The solution to $Ax = 0$ is given by $\operatorname{span} \{ q_3, q_4 \}$ (because $q_3,q_4$ are orthogonal to $q_1,q_2$) and so the solution to $Ax = z_1$ is given by
$$q_1 + \operatorname{span} \{ q_3, q_4 \}.$$
• $q_1,..,q_4$ are orthonormal so If you want just edit the solution to $q_1 + span\{q_3,q_4\}$ – Dragan Zrilić Jun 12 '17 at 7:55