Interpretation of the null space of a 9x9 matrix

Question

I have a system of equations $A_{9\times 9}x_{9\times 1}=b_{9\times 1}$, where the right hand side has 3 unknowns (an underdetermined system). $\lambda_i$ are unknown and $L_i$ are known.

$$A\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7\\x_8\\x_9\end{bmatrix} = \begin{bmatrix}\lambda_1L_1\\\lambda_2L_1\\\lambda_3L_1\\\lambda_1L_2\\\lambda_2L_2\\\lambda_3L_2\\\lambda_1L_3\\\lambda_2L_3\\\lambda_3L_3\end{bmatrix}$$

I also know that $\lambda_1^2 + \lambda_2^2 + \lambda_3^2=1$. I thought at least I should be able learn something about the solution from the null space of $A$ and this one extra equation for $\lambda_i$.

If:

>> A = [ 383.54       1056.74      -1894.76             0             0             0             0             0          0
              0             0             0        383.54       1056.74      -1894.76             0             0             0
              0             0             0             0             0             0        383.54       1056.74      -1894.76
         383.54       1056.74      -1894.76             0             0             0             0             0             0
              0             0             0        383.54       1056.74      -1894.76             0             0             0
              0             0             0             0             0             0        383.54       1056.74      -1894.76
         514.64       1479.24      -2693.77             0             0             0             0             0             0
              0             0             0        514.64       1479.24      -2693.77             0             0             0
              0             0             0             0             0             0        514.64       1479.24      -2693.77])


>> null(A) % Null sapce of A
     -0.01          0.57          0.05
      0.02         -0.76         -0.07
      0.01         -0.31         -0.03
      0.57          0.02         -0.04
     -0.76         -0.02          0.06
     -0.31         -0.01          0.02
     -0.05          0.05         -0.57
      0.06         -0.07          0.75
      0.02         -0.03          0.31

Could someone kindly help me what the interpretation of the null space of $A$, for this particular example, is? How can I use it to have a sense of what the solution of $Ax = b$ can be?

Answer 1

I'm not sure what you mean by interpretation of the null space. The null space is the set(space) of vectors mapping to the null vector under a linear transformation. Indeed this set always forms a subspace of the domain space. So in terms of maybe a geometric interpretation(if adequate, that is if there is such a natural interpretation of the underlying vector space like in $\mathbb{R}^n$, which I try to focus on), the null space may range from being a line to being a hyperplane in $n$-dimensional space, depending on its dimension. (including degenerate cases like the space consisting just of the null-vector and the whole space) Nevertheless, I now just recall its definition, supposing a classical real setting and with $A\in\mathbb{R}^{(n,n)}$:

$N(A)=\{x\in\mathbb{R}^n\mid Ax=\mathbf{0}\}$

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The peculiar property of the null space of a coefficient matrix corresponding to a linear system of equations, given over $Ax=b$, is that for any already found solution $x$, the whole (affine) solution space is given by $x+N(T)=\{x+v\mid v\in N(T)\}$. That is, the solution space of a general equation is always a translation of the null space by some(any!) solution to the equation. In other words, knowing the null space also tells you about many basic and important properties for general solution spaces(if non-empty), like e.g. dimension etc.

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To see this formally, think about $S(A,b)=\{x\in\mathbb{R}^n\mid Ax=b\}$. Supposing that this set is non-empty, let $v\in S(A,b)$. We show $v+N(A)=S(A,b)$ via double inclusion:

Let $x\in v+N(A)$, i.e. $x=v+w$ for $w\in N(A)$. Now, $Ax=A(v+w)=Av+Aw=Av=b$, as $w\in N(A)$. Thus $x\in S(A,b)$.

Let $x\in S(A,b)$, i.e. $Ax=b$. Thus $x-v\in N(A)$ as $A(x-v)=Ax-Av=b-b=\mathbf{0}$. But now $v+(x-v)=x\in v+N(A)$.

Answer 2

Let me expand my comment in a fashion which is perhaps more understandable; your system $Ax=b$ is equivalent to $$ \underbrace{\left[ \begin{array}{cccc} A_1 & -L_1I_3 \\ A_2 & -L_2I_3 \\ A_3 & -L_3I_3 \end{array} \right]}_{B}\left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \\ x_7 \\ x_8 \\ x_9 \\ \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \end{array}\right] = 0$$ Where $A_1$ is the submatrix of $A$ formed by its first three rows, $A_2$ is the submatrix formed by the middle three rows and $A_3$ is the last three rows in $A$. The notation $I_3$ is for the $3 \times 3$ identity matrix. Then since all the entries in the $9 \times 12$ matrix $B$ are given you can calculate the null space of $B$ and impose your constraint $\lambda_1^2+\lambda_2^2+\lambda_3^2=1$ on the resulting solution. Unless of course the solution does not allow such $\lambda$... which I would not be able to guess without running this through some software.