# Determine $N(A),R(A),R(A^T)$ in terms of eigenvectors

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.

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 $$a_1 \neq 0$$.

• Thanks for answering. Is that true for all matrices that column space can be spanned by eigenbasis? – techno Jun 14 '19 at 17:57
• $R(A^T)$ is indeed the row space of $A$. The column space of $A$ is $\text{im}(A)$. Aside from that it's a good answer... – Christiaan Hattingh Jun 14 '19 at 18:42
• 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. – paulinho Jun 14 '19 at 18:58
• @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 $C(A)$. – paulinho Jun 14 '19 at 19:00
• 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. – 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...
• 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. :) – techno Jun 14 '19 at 18:14
• 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 – Christiaan Hattingh Jun 14 '19 at 18:20
• 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! – techno Jun 14 '19 at 18:31