0
$\begingroup$

I have question regarding fundamental subspaces and eigenvectors.

Problem:

Let $0,1,2$ be eigenvalues with eigenvectors $x_1,x_2,x_3$, respectivly, of matrix $A$.

Determine kernel, image and $R(A^T)$ in terms of $x_1,x_2,x_3$

Determine all solutions of system $Ax=-2x_2-3x_3$ in terms of $x_1,x_2,x_3$.

What you can say about system $Ax=x_1+x_3$?

Is $A$ orthogonal matrix?

I tried to solve it. Here is few points I have:

The subspace of $x_1$ (with zero vector) is $\text{ker}(A)$. The other two make up $R(A)$.

Really don't know how to represent $Ax=-2x_2-3x_3$ (this part seems hardest to me) $Ax=x_1+x_3\longrightarrow Ax=x_3$ because $x_1$ (including zero vector) represents $\text{ker}(A)$.

$A$ is not orthogonal matrix, because: the lengths of eigenvalues are not $1$, we don't know are the eigenvectors are orthogonal to each other.

Hopefully someone can give me few hints on this. I tried to go through books, get more theory but I didn't find anything that connects eigenvalues and eigenvectors directly to fundamental subspaces besides the kernel. Thank you in advance.

$\endgroup$
0
$\begingroup$

I'm going to assume you're dealing with a $3\times3$ matrix here. The key here is to recognize that eigenvectors must be linearly independent if they correspond to different eigenvalues. The column space of $A$ is spanned by the nonzero eigenvectors as this matrix has an eigenbasis (a set of eigenvectors that span $\mathbb{R}^3)$: since any vector $\vec{v} \in \mathbb{R}^3$ can be written as a unique linear combination of the eigenvectors, the space $A\vec{x} = A (a_1 \vec{x}_1 + a_2 \vec{x}_2 + a_3\vec{x}_3) = a_2 \vec{x}_2 + 2a_3\vec{x}_3$, for any $a_2, a_3 \in \mathbb{R}$. To find the row space, look for the subspace of $\mathbb{R}^3$ that is perpendicular to the null space. So the row space of $A$ here is $\{\vec{v} \text{ }| \text{ } \vec{v} \cdot \vec{x}_1 = 0\}$.

Again, we know that $A\vec{x}$ can be rewritten as $A(a_1 \vec{x}_2 + a_2 \vec{x}_2 + a_3\vec{x}_3) = a_2 \vec{x}_2 + 2a_3\vec{x}_3$, because all the $\vec{x}_i$ are independent and thus form a basis for $\mathbb{R}^3$. Equating the two sides, we must have $a_2 = -2, a_3 = -3/2$. However, we can choose any value of $a_1 \in \mathbb{R}$ because $A$ maps any multiple of $\vec{x}_1$ to the zero vector. So any vector $\vec{v}$ in the form $\vec{v} = a_1 \vec{x}_1 - 2\vec{x}_2 - \frac{3}{2}\vec{x}_3$ is a solution to this equation.

We know that any linear combination including $\vec{x}_1$ is inconsistent, because the solution to $A \vec{x} = \vec{b}$ is spanned solely by vectors $\vec{x}_2$ and $\vec{x}_3$. Because $\vec{x}_1, \vec{x}_2$, and $\vec{x}_3$ are linearly independent, we know there is no such solution to $A \vec{x} = a_1 \vec{x}_1 + a_2 \vec{x}_2 + a_3 \vec{x}_3$ with $a_1 \neq 0$.

$\endgroup$
  • $\begingroup$ Thanks for answering. Is that true for all matrices that column space can be spanned by eigenbasis? $\endgroup$ – techno Jun 14 '19 at 17:57
  • $\begingroup$ $R(A^T)$ is indeed the row space of $A$. The column space of $A$ is $\text{im}(A)$. Aside from that it's a good answer... $\endgroup$ – Christiaan Hattingh Jun 14 '19 at 18:42
  • $\begingroup$ Actually, it isn't true that all matrices have an eigenbasis! In fact, a matrix has an eigenbasis if and only if it is similar to a diagonal matrix (of its eigenvalues). Such matrices are called deficient. $\endgroup$ – paulinho Jun 14 '19 at 18:58
  • $\begingroup$ @ChristiaanHattingh Unless I've completely misunderstood the notation, $R(A)$ is the row space of matrix $A$. So $R(A^T$) is the row space of the transpose matrix $A^T$, which is equivalent to $C(A)$. $\endgroup$ – paulinho Jun 14 '19 at 19:00
  • $\begingroup$ Hi @paulinho...no usually $R(A)$ indicates the range of $A$, at least in all the articles and texts I have encountered...in this context it seems to be the intention too. But agreed, if it's not defined properly then it can be confusing. $\endgroup$ – Christiaan Hattingh Jun 14 '19 at 20:16
0
$\begingroup$

As pointed out by @paulinho it is crucial to know that $A$ is $3 \times 3$, else the spaces cannot be fully described in terms of $x_1, x_2, x_3$. I will add some further notes/hints, assuming $A$ is $3 \times 3$.

  1. Since the $R(A)$ is spanned by $x_2$ and $x_3$, which is linearly independent from $x_1$, is it possible to find $x$ such that $Ax=x_1+x_3$?
  2. Can a singular matrix be orthogonal? - I think this is the key to the last question...
$\endgroup$
  • $\begingroup$ Hi, thanks for giving me hints. 1) I assume it will be A(ax1+x3/2) 2) Orthogonal matrix determinant is 1 or -1 and ours is zero because the rank is not full. :) $\endgroup$ – techno Jun 14 '19 at 18:14
  • $\begingroup$ for 1) you then have $A(ax_1+x_3/2)=x_3 \neq x_1 + x_3$ ... so the point is you cannot find $x$ that satisfies this equation $\endgroup$ – Christiaan Hattingh Jun 14 '19 at 18:20
  • $\begingroup$ Oooh now i get it. x1 is in the ker(A) so in any way it can't be in im(A). I supose 2) is correct. Thanks! $\endgroup$ – techno Jun 14 '19 at 18:31

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.