I have question regarding fundamental subspaces and eigenvectors.


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.


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 $a_1 \neq 0$.

  • $\begingroup$ 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

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...
  • $\begingroup$ 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

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