My question differs from this question. Source for the following is on P3 hereof:

Suppose $A$ has eigenvalues $0,3,5$ with linearly independent eigenvectors $u,v,w$.
(a) Give a basis for $\operatorname{null}(A)$ and a basis for $\operatorname{col}(A)$.
(b) Show that $Ax=u$ has no solution.
Hint: If it did, then $()$ would be in $\operatorname{col}(A)$ and this contradicts the assumption.


(a) $\operatorname{null}(A)=\operatorname{null}(A-0I)=E_0=\operatorname{span}(u)$. For any linear combination $c_1v+c_2w$,
$c_1\color{green}{v}+c_2\color{#FF4F00}w = c_1\color{green}{\dfrac{Av}{3}} + c_2\color{#FF4F00}{\dfrac{Aw}{5}} = A\left(\dfrac{c_1}3 v + \dfrac{c_2}5 w\right)\in\operatorname{col}(A),$
therefore $\operatorname{col}(A)=\operatorname{span}(v,w)$.
(b) $Ax=\color{green}v+\color{#FF4F00}w =\color{green}{\frac 1 3 A v}+\color{#FF4F00}{\frac 1 5 Aw} =A\left(\dfrac v 3 + \dfrac w 5\right)$.
All solutions are of the form $\dfrac v 3 + \dfrac w 5 + cu$.
(c) Assume that $Ax=u$ has a solution. Then $u\in\operatorname{col}(A)$,
but $u$ is linearly independent of both $v$ and $w$ therefore cannot be in $\operatorname{col}(A)$.

By P308, Thm 4.20, from David Poole's Linear Algebra, distinct eigenvalues correspond to linearly independent eigenvectors. Suppose I revamp part (c)'s argument for $\mathbf{v}$ instead:

Assume $\mathbf{x}$ solves $\mathbf{Ax = v}$. Then $\mathbf{v} \in col(A)$.
By Thm 4.20 (aforementioned), $\{\mathbf{u, v, w}\}$ is lin-ind therefore $\mathbf{v} \notin col(A)$.

$1.$ This is false because $\color{green}{\mathbf{Av = 3v}}$ is given. Thus, what am I misconstruing?
Why does the argument in part (c) hold for $\mathbf{u}$ but fail for $\mathbf{v}$ and $\mathbf{w}$ ?

$2.$ Moreover, what's the intuition for part (b)? P267, 268 of Poole presents the geometric interpretation of eigenvectors so I'd imagine something geometric here?

$\Large{\text{ Supplementary dated Jan 12 2014: }}$

$3.$ In (a), how would you divine/previse to consider $c_1\color{green}{v}+c_2\color{#FF4F00}w$ for determining $colspace(A)$?

$4.$ As regards (b), what's the objective in solving $Ax=\color{green}v+\color{#FF4F00}w$ for $x$? Why be concerned with this?

  • 1
    $\begingroup$ Both your question and the link you gave do not mention, surprisingly, what is $\;A\;$, but it seems to be a $\;3\times 3\;$ matrix. Is this correct? And why your solutions has (a)-(b)-(c) if the question only has (a)-(b)? $\endgroup$ – DonAntonio Jan 3 '14 at 14:11
  • $\begingroup$ If I understood correctly what's going on here, and that could be a long shot, the reasoning seems to be simply that $\;u\notin Span\,\{v\,,\,w\}=Col(A)\;$ so no solution's possible, but for $\;w,v\;$ there're trivially solutions... $\endgroup$ – DonAntonio Jan 3 '14 at 14:16

Let's address your question 1 first. Recall that in part a you found that $\{\mathbf{v},\ \mathbf{w}\}$ forms a basis for the columnspace of $A$. The reason that the argument works for $\mathbf{u}$ and not for $\mathbf{v}$ is because $\mathbf{u}$ is not an element of this basis set while $\mathbf{v}$ is. Analagously, the argument will also fail for $\mathbf{w}$ (and indeed any linear combination of $\mathbf{v}$ and $\mathbf{w}$).

Since $\{\mathbf{u},\ \mathbf{v},\ \mathbf{w}\}$ was assumed to be linearly independent, it follows that $\mathbf{u}$ is not in the span of $\mathbf{v}$ and $\mathbf{w}$. Otherwise, there exists scalars $a$ and $b$ such that $$a\mathbf{v} + b\mathbf{w} = \mathbf{u}$$ But then we have the non-trivial linear combination $$a\mathbf{v} + b\mathbf{w} - \mathbf{u} = \mathbf{0}$$ contrary to the assumption of linear independence.

Since the span of $\mathbf{v}$ and $\mathbf{w}$ is precisely $\mathrm{col}(A)$, this means that $\mathbf{u}$ is not an element of the columnspace, i.e. there does not exist $\mathbf{x}$ such that $A\mathbf{x} = \mathbf{u}$. Note that this argument cannot work for $\mathbf{v}$ because $\mathbf{v}$ is in the span of $\{\mathbf{v},\ \mathbf{w}\}$, and trivially so.

For question 2, the geometric intuition is complicated by the fact that there do exist matrices in which vectors in the nullspace are also in the columspace (i.e. matrices for which part b has a solution). In fact, there exists matrices in which the columnspace is equal to the nullspace, for example $$A = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$$ However, for diagonalizable matrices (such as the one in consideration), we can present a relatively simple view. If you have not learned about diagonalization yet, then it may be difficult to follow the next bit. I recommend that you continue with your studies for now; diagonalization is typically not far away once you've begun studying eigenvectors and eigenvalues. You can come back to the next part after you've learned about diagonalization.

When your matrix is diagonalizable, you can effectively view the matrix as a sequence of scalings along the standard axes (via a change of basis). With this viewpoint, we can effectively look at $A$ as a linear transformation in 3-space where the $x$-axis is stretched by a factor of $3$ (representing the basis vector $\mathbf{v}$), where the $y$-axis is stretched by a factor of $5$ (representing $\mathbf{w}$), and where the $z$-axis is compressed (representing $\mathbf{u}$).

From this viewpoint, the statement that $\mathbf{u}$ is not in the columnspace is analagous to the statement that the image of the linear transform does not contain the $z$-axis, but this is intuitively obvious since the mapping compresses $z$. In fact, it's relatively clear from our description that the image of the map is entirely contained within the $z=0$ plane, i.e. the $xy$-plane. This is a representation of the fact that the columnspace of $A$ does not contain any vectors in which $\mathbf{u}$ is a component, i.e. $a\mathbf{v} + b\mathbf{w} + c\mathbf{u}$ is not an element of $\mathrm{col}(A)$ if $c\neq 0$.

Addendum for the supplementary questions by the OP

Question 3: Why consider $c_1\mathbf{v} + c_2\mathbf{w}$ for the columnspace?

The consideration of linear combinations of $\mathbf{v}$ and $\mathbf{w}$ comes from experience with eigenvalues. The key point is the hidden assumption that the matrix is $3\times 3$ which is not stated in the text of the question for some reason. Since $A$ is $3\times 3$ with $3$ distinct eigenvalues, this means that $A$ is diagonalizable, as I mentioned previously. What this means in particular is that the eigenvectors $\mathbf{u},\ \mathbf{v}$ and $\mathbf{w}$ forms a basis for $\mathbb{R}^3$ (assuming $A$ is a real matrix). If we consider some arbitary vector in $\mathbb{R}^3$, then it follows that we can write $\mathbf{x}$ as a linear combination of our basis of eigenvectors or eigenbasis: $$\mathbf{x} = c_1\mathbf{v} + c_2\mathbf{w} + c_3\mathbf{u}$$ If we then consider the image of $\mathbf{x}$ under $A$, we get $$A\mathbf{x} = c_1A\mathbf{v} + c_2A\mathbf{w} + c_3A\mathbf{u} = 3c_1\mathbf{v} + 5c_2\mathbf{w}$$ Since our choice of $\mathbf{x}$ is completely arbitrary, it follows that the entire image of $A$, that is the entire columnspace of $A$, is expressible as a linear combination of $\mathbf{v}$ and $\mathbf{w}$. Conversely, the solution for part (a) shows that every such linear combination is indeed in the columnspace.

As for question 4, I'm also not too sure what the solution is trying to do. It seems that part (b) in the question is answered by part (c) in the solution and that part (b) of the solution is completely extraneous. It seems to answer a different question altogether.

Addendum 2 in response to OP's questions

The fact that $\mathbf{u}\notin \mathrm{span}(\mathbf{v},\ \mathbf{w})$ comes simply from the fact that $\{\mathbf{u},\ \mathbf{v},\ \mathbf{w}\}$ is a linearly independent set. Certainly you can recast the argument so that contradiction is not used, and here is one example of a proof using the contrapositive instead of contradiction. This proof makes it clear that the result follows essentially from the definition of linear independence.

Theorem: Suppose that $\{\mathbf{v}_1,\ \cdots,\ \mathbf{v}_n\}$ is a linearly independent set of vectors. Then $\mathbf{v}_1 \notin \mathrm{span}(\mathbf{v}_2,\ \cdots,\ \mathbf{v}_n)$.

Proof: The contrapositive of the statement is: Suppose that $\mathbf{v}_1 \in \mathrm{span}(\mathbf{v}_2,\ \cdots,\ \mathbf{v}_n)$. Then $\{\mathbf{v}_1,\ \cdots,\ \mathbf{v}_n\}$ is linearly dependent.

But this is essentially trivial. The fact that $\mathbf{v}_1 \in \mathrm{span}(\mathbf{v}_2,\ \cdots,\ \mathbf{v}_n)$ means there exists some linear combination of $\{\mathbf{v}_2,\ \cdots,\ \mathbf{v}_n\}$ adding up to $\mathbf{v}_1$, i.e. there exists scalars $\{c_i\}$ such that $$\mathbf{v}_1 = c_2\mathbf{v}_2 + \cdots + c_n\mathbf{v}_n$$ But then we also have $$-\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_n\mathbf{v}_n = \mathbf{0}$$ which is a non-trivial linear combination of $\{\mathbf{v}_1,\ \cdots,\ \mathbf{v}_n\}$ adding up to the zero vector. By definition, this means that $\{\mathbf{v}_1,\ \cdots,\ \mathbf{v}_n\}$ is linearly dependent. $\square$

Applying this result to the set $\{\mathbf{u},\ \mathbf{v},\ \mathbf{w}\}$ gives the desired result.

Note that the above argument works for any set of linearly independent vectors. However, we can also take advantage of the fact that $\{\mathbf{u},\ \mathbf{v},\ \mathbf{w}\}$ are eigenvectors under distinct eigenvalues to prove that they are linearly independent. This results in the well known theorem that eigenvectors corresponding to distinct eigenvalues are linearly independent. The proof of this fact is available in many, many locations and so I will not reproduce it here. I simply wanted to let you know of an alternative approach which you may prefer instead.

  • $\begingroup$ +1. Thank you very much for your sterling detailed and helpful answer. I'd be grateful for your superlative attention and help towards my supplementary questions in my OP. If possible, would you please answer in your answer and not as a comment? $\endgroup$ – Greek - Area 51 Proposal Jan 12 '14 at 12:51
  • $\begingroup$ @LePressentiment I've tried to answer your supplementary questions. Next time, please consider starting a new question and simply referencing the old. As for your question 4, I am also quite confused as to what the solution is trying to do. It seems extraneous to me. $\endgroup$ – EuYu Jan 13 '14 at 1:05
  • $\begingroup$ Thank you profoundly again! Did you mean to write $\mathbf{v, w}$ where I inserted red brackets? Please feel free to revert the edit. Lastly, I apprehend your 2nd paragraph but is it possible to intuit $u \notin span\{v, w\}$ without contradiction? Would you please recast it as a direct natural argument? $\endgroup$ – Greek - Area 51 Proposal Jan 13 '14 at 15:17
  • $\begingroup$ @LePressentiment I did mean to write $\mathbf{v}$ and $\mathbf{w}$, thanks for catching that typo. I've also added a section addressing your new question. $\endgroup$ – EuYu Jan 14 '14 at 5:59
  • $\begingroup$ Thank you profoundly for your continual care. Please maintain your supernal contributions and beneficence, which I'm grateful for and cherish. $\endgroup$ – Greek - Area 51 Proposal Jan 25 '14 at 6:19

You have the right ideas, but you're expressing them in a very tortuous way.

If the vectors $u$, $v$ and $w$ are linearly independent, they must be relative to distinct eigenvalues (assuming $A$ is $3\times3$). I believe it's also intended that $Au=0$, $Av=3v$ and $Aw=5w$, but in this case linear independence would follow.

Moreover $\{u,v,w\}$ is a basis of $\mathbb{R}^3$ (or $\mathbb{C}^3$, if you use complex numbers).

Each eigenspace has dimension $1$, so $u$ generates the null space.

An equation $Ax=y$ has a solution if and only if $y$ belongs to the column space of $A$; but, since $v$ and $w$ obviously belong to the column space, they form a basis of it, because of the rank-nullity theorem. It follows that $u$ doesn't belong to the column space.

Note that saying $u$ is linearly independent from both $v$ and $w$ is meaningless.

  • $\begingroup$ +1. Thank you very much for your answer. Would you please expound on "$u$ is linearly independent from both $v$ and $w$ is meaningless"? What should the author have written? I'd have written $\{u, v, w\}$ is linearly independent. I'd also be grateful for your help towards my supplementary in my OP. $\endgroup$ – Greek - Area 51 Proposal Jan 12 '14 at 12:52
  • $\begingroup$ @LePressentiment The statement is too ambiguous to have a sensible meaning. $\endgroup$ – egreg Jan 12 '14 at 13:42
  • $\begingroup$ I don't perceive what you mean. Would you please enlarge on your comment in your answer? Please forgive me for forgetting +1. $\endgroup$ – Greek - Area 51 Proposal Jan 13 '14 at 15:18

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