Dimension of all four fundamental subspaces of a matrix.

Question

For the system of linear equations $Ax = b$ with $b =\begin{bmatrix} 4\\ 6\\ 10\\ 14 \end{bmatrix}\\ $. The set of solutions is given by- $\left\{ x : x = \begin{bmatrix} 0\\ 0\\ -2 \end{bmatrix} + c \begin{bmatrix} 0\\ 1\\ 0\end{bmatrix} + d \begin{bmatrix} 1\\ 0\\ 1\end{bmatrix} \right\}$.

The question requires to find the matrix $A$ and dimensions of all four fundamental subspaces of $A$. Is there any intuitive way of finding the matrix $A$, since the remaining problem becomes straightforward thereafter. Thanks in advance for any help.

Answer 1

The matrix $A$ is of the form$$A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\a_{41}&a_{42}&a_{43}\end{bmatrix}.$$But you are told that $A.(0,0,-2)=(4,6,10,14)$; in other words$$A=\begin{bmatrix}a_{11}&a_{12}&-2\\a_{21}&a_{22}&-3\\a_{31}&a_{32}&-5\\a_{41}&a_{42}&-7\end{bmatrix}.$$Also, it follows from your assumptions that $A.(0,1,0)=(0,0,0,0)$ that$$A=\begin{bmatrix}a_{11}&0&-2\\a_{21}&0&-3\\a_{31}&0&-5\\a_{41}&0&-7\end{bmatrix}.$$Finally, $A.(1,0,1)$ is also $(0,0,0,0)$. So$$A=\begin{bmatrix}2&0&-2\\3&0&-3\\5&0&-5\\7&0&-7\end{bmatrix}.$$Can you take it from here?

Answer 2

It's not actually necessary to find the matrix. Here's a way to approach the problem.

• From the shape of $b$, we know that the output of the transformation $T(x) - Ax$ is an element of $\Bbb R^4$. This tells us that $A$ has $4$ rows
• From the shape of the vectors in the solution set, we know that the inputs to $T$ are elements of $\Bbb R^3$. This tells us that $A$ has $3$ columns.
• Recall that there is a connection between the nullspace of $A$ and the solution space to $Ax = b$. What is the dimension of the nullspace of $A$?
• Once you have the nullspace of $A$, the rank-nullity theorem (Strang might call it the "dimension theorem") gives you the rank of $A$, which gives you the dimension of two more fundamental subsapaces.
• To get the dimension of the last fundamental subspace, use the rank-nullity theorem again.