Find one matrix $A\in M_3,_4(\mathbb R)$ such that $N(A)=L((-3,1,0,0)(-2,0,-6,1))$ then decribe all that matrices.

We need to find solutin for $Ax=0$ where $x$ can show as linear combination of $(-3,1,0,0)$ and $(-2,0,-6,1)$

I find this matrices

A=$\begin{bmatrix} 1& 3& 1& 8\\ 0& 0& 1& 6\\ 0& 0& 0& 0 \end{bmatrix}$.

but I need to decribe all matrices, this matrices look after some elementary transformation. But from here we can see that third row can show as linear combination of other two $A_{3\cdot}=\gamma A_{2\cdot}+\omega A_{1\cdot}$, $A_{2\cdot}=\begin{bmatrix} \alpha a& \beta b& e& f \end{bmatrix}$. $A_{1\cdot}=\begin{bmatrix} a& b& c& d \end{bmatrix}$.

A=$\begin{bmatrix} a& b& c& d\\ \alpha a& \beta b& e& f\\ \gamma a+\alpha\omega a& \gamma b+\beta b& \gamma c+\omega e& \gamma d+\omega f \end{bmatrix}$. $a\not=0$

something like that but I need yours opinion, what you think?

  • $\begingroup$ This doesn’t have anything to do with eigenvalues. $\endgroup$ – amd Sep 3 '18 at 21:39

Take the standard basis. Now find the unit vectors linearly independent of the given vectors say {$e\space, f$} .Since $\space N(A) =2$ then rank of $A$ is 2 by Dimension Theorem. Now By Gram–Schmidt orthogonalization process find the basis of $N(A) ^\perp$

  • $\begingroup$ The rank of $A$ is 2, though. It will have 2 pivotal columns. $\endgroup$ – Arthur Sep 3 '18 at 6:09
  • $\begingroup$ yes now i see that rank A=2 because this matrix can be of some linear operation $\mathbb R^4\to \mathbb R^3$ $\endgroup$ – Marko Škorić Sep 3 '18 at 6:12
  • $\begingroup$ Yeah Sorry... MY Mistake $\endgroup$ – Sadil Khan Sep 3 '18 at 6:21
  • $\begingroup$ This doesn’t describe all such matrices. For example, the matrix with rows $2e$, $0$, $2f$ also has the correct null space, but doesn’t fit your description in several ways. Also, you can’t just take any vectors that are not elements of $N(A)$: you need a basis for $N(A)^\perp$. $\endgroup$ – amd Sep 3 '18 at 21:47
  • $\begingroup$ Your example also has 2 pivotal columns $\endgroup$ – Sadil Khan Sep 3 '18 at 22:05

The row space of a matrix is the orthogonal complement of its null space. Since $\dim N(A)=2$, the dimension of the row space is also $2$, therefore the matrices that satisfy the given conditions have two linearly-independent rows that span $N(A)^{\perp}$, with the remaining row a linear combination of the other two.

You’ve already found a basis for $N(A)^\perp$, so we can say that each row of $A$ must be a linear combination of $(1,3,1,8)$ and $(0,0,1,6)$, subject to the additional conditions described above.

Another way to characterize this set of matrices is that they are of the form $BA$, where $A$ is the matrix that you’ve already found and $B$ is any invertible $3\times3$ matrix. The rows of the product $BA$ are all linear combinations of the rows of $A$, and requiring that $B$ be invertible guarantees that the null space won’t change.


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