How to construct a matrix given the null basis of A?

Question

Construct a $4\times4$ matrix $A$ such that $\{(1,2,3,4),(1,1,2,2)\}$ is a basis of $N(A)$.

So I know that $A$ will have two pivot columns and two free columns, but beyond this I'm not sure how to approach/solve.

Answer 1

Here is a more detailed tip based on zzuussee's comment:

Let

$$A=\begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\\a_{41}&a_{42}&a_{43}&a_{44}\end{bmatrix}.$$

What could we say about each $a_{ij}$ if we require that

$$A\begin{bmatrix}1\\2\\3\\4\end{bmatrix}=A\begin{bmatrix}1\\1\\2\\2\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}?$$

Can you choose values for each $a_{ij}$ that satisfy the above requirements?

Answer 2

Let me show you how it would go. Call $v_1=(1,2,3,4)$ and $v_2=(1,1,2,2)$.

Let $v=(x_1,x_2,x_3,x_4)$. We want $v\cdot v_1=0$ and $v\cdot v_2=0$. This gives us the system of equations $$\begin{align}x_1+2x_2+3x_3+4x_4&=0\\x_1+x_2+2x_3+2x_4&=0\end{align}$$

Subtracting the first equation from the second we get: $$\begin{align}x_1+2x_2+3x_3+4x_4&=0\\\phantom{x_1}-x_2-x_3-2x_4&=0\end{align}$$

Now the system is in echelon form. In order to get two linearly independent solutions we can put first $x_3=1$, $x_4=0$ and solve for $x_1,x_2$, and then put $x_3=0$, $x_4=1$ and solve for $x_1,x_2$.

Doing that gives us two vectors $$\begin{align}u_1&=(-1,-1,1,0)\\u_2&=(0,-2,0,1)\end{align}$$

Therefore, one matrix satisfying the conditions of the problem would be $$\begin{pmatrix}-1,&-1,&1,&0\\0,&-2,&0,&1\\-1,&-1,&1,&0\\0,&-2,&0,&1\end{pmatrix}$$

obtained by putting $u_1,u_2$ and again $u_1,u_2$ as rows.

Answer 3

It's well-known that a matrix is determined by its effect on a basis. So let's complete our null basis to a basis for $\mathbb R^4$. For instance, we could use $e_1=(1,0,0,0)$ and $e_2=(0,1,0,0)$.

Now, if we call our basis $\beta$, then $[T]_{\beta}^{\beta}=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$ would be a matrix with the right null space (actually the first two columns could be any two vectors that are linearly independent...)

Meanwhile, the change of basis matrix from $\beta$ to the standard basis is: $P=\begin{pmatrix}1&0&1&1\\0&1&2&1\\0&0&3&2\\0&0&4&2\end{pmatrix}$.

So, we could let $A=P[T]_{\beta}^{\beta}P^{-1}$.

I get $A=\begin{pmatrix}1&0&-1&\frac12\\0&1&0&-\frac12\\0&0&0&0\\0&0&0&0\end{pmatrix}$.

Answer 4

The row space of a matrix is the orthogonal complement of its null space. So, you can construct the required matrix by finding a basis for this orthogonal complement. In this case, this will give you two of the rows, and the other two rows can be any linear combinations of those two rows, including rows of all zeros.

Calling the two given vectors $\mathbf n_1$ and $\mathbf n_2$, the orthogonal complement of their span is the set of all vectors $\mathbf x$ that satisfy $\mathbf n_1\cdot\mathbf x=\mathbf n_2\cdot\mathbf x=0$. This is a pair of homogeneous linear equations in the components of $\mathbf x$, so $\mathscr N(A)^\perp$ is the null space of the matrix $\small{\begin{bmatrix}\mathbf n_1 & \mathbf n_2\end{bmatrix}}^T$. I’m sure you know how to compute that.

Answer 5

You could for instance say that you have a matrix \begin{equation} A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \\ \end{bmatrix} \end{equation} such that \begin{equation} AX = 0 \end{equation} where \begin{equation} X = \begin{bmatrix} 1 & 1 \\ 2 & 1 \\ 3 & 2\\ 4 & 2 \end{bmatrix} \end{equation} Hence we have to find $a_{ij}$'s such that \begin{equation} \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \\ \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & 1 \\ 3 & 2\\ 4 & 2 \end{bmatrix} = 0 \end{equation} i.e. we have the following system to solve: \begin{align} a_{11} + 2a_{12} + 3a_{13} + 4a_{14} & = 0 \\ a_{21} + 2a_{22} + 3a_{23} + 4a_{24} & = 0 \\ a_{31} + 2a_{32} + 3a_{33} + 4a_{34} & = 0 \\ a_{41} + 2a_{42} + 3a_{43} + 4a_{44} & = 0 \\ a_{11} + a_{12} + 2a_{13} + 2a_{14} & = 0 \\ a_{21} + a_{22} + 2a_{23} + 2a_{24} & = 0 \\ a_{31} + a_{32} + 2a_{33} + 2a_{34} & = 0 \\ a_{41} + a_{42} + 2a_{43} + 2a_{44} & = 0 \end{align} This is exhaustive as it is a system of 8 equations in 16 unknowns. Therefore, we will have infinitely many solutions. \\ For example, you could construct a projector matrix $P_X$ that spans the columns of $X$, i.e. \begin{equation} P_X = X(X^TX)^{-1}X^T = \begin{bmatrix} 0.5455 & -0.0909 & 0.4545 & -0.1818\\ -0.0909 & 0.1818 & 0.0909 & 0.3636\\ 0.4545 & 0.0909 & 0.5455 & 0.1818\\ -0.1818 & 0.3636 & 0.1818 & 0.7273 \end{bmatrix} \end{equation} And then you can say that my matrix $A$ spans the null space of $P_X$, i.e. \begin{equation} A = I - P_X = \begin{bmatrix} 0.4545 & 0.0909 & -0.4545 & 0.1818\\ 0.0909 & 0.8182 & -0.0909 & -0.3636\\ -0.4545 & -0.0909 & 0.4545& -0.1818\\ 0.1818 & -0.3636 & -0.1818 & 0.2727 \end{bmatrix} \end{equation} Now check,

\begin{equation} AX =(I - P_X)X = X - P_XX = X - X(X^TX)^{-1}X^TX = X-X = 0 \end{equation} and voila there you have a matrix with nullspace being the columns of $X$.