1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ This doesn’t have anything to do with eigenvalues. $\endgroup$ – amd Sep 3 '18 at 21:39
0
$\begingroup$

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 $N(A) ^\perp$

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.