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let $(e_1,e_2,...,e_5)$ canonical basis of $R^5$, $V=(a,b,c,d,e)\in R^5$ with $V\neq(0,0,0,0,0)$. we consider $f:R^5\to R^5$ and its matrix :

$$Mat(f) = M= \begin{pmatrix} a&a&a&a&a\\ b&b&b&b&b\\ c&c&c&c&c\\ d&d&d&d&d\\ e&e&e&e&e\\ \end{pmatrix}$$

It's rank is $rg(f)=1$ because $V\neq(0,0,0,0,0)$ and all columns are the same. So $Dim(Im(f))=1$ and because of the rank theorem, $Dim(Ker(f))=4$

Basis of $Im(f)=Vect\begin{pmatrix}\begin{pmatrix}a\\b\\c\\d\\e\\\end{pmatrix}\end{pmatrix}$ and by solving with $X=\begin{pmatrix}x\\y\\z\\t\\u\\ \end{pmatrix}$, $ MX=0$ I find : $Ker(M)=Vect\begin{pmatrix}\begin{pmatrix}-1\\1\\0\\0\\0\end{pmatrix},\begin{pmatrix}-1\\0\\1\\0\\0\\\end{pmatrix},\begin{pmatrix}-1\\0\\0\\1\\0\end{pmatrix},\begin{pmatrix}-1\\0\\0\\0\\1\end{pmatrix}\end{pmatrix}$

I was then asked a condition on $a,b,c,d,e$ for $V\in Ker(f)$ and I found $a+b+c+d+e=0$ by solving $MV=0$ And next is a question I need to verify:

Question : let this condition be fullfilled, let $(\epsilon_1,\epsilon_2,\epsilon_3,\epsilon_4)$ basis of $Ker(f)$ chosen so $\epsilon_4=V$ Complete it in a basis of E (here $R^5$) and write the matrix in the new basis

Verification : what I did : Is it okay for me to set $V = \begin{pmatrix}-1\\0\\0\\0\\1\end{pmatrix}$ so it would mean that my matrix $M$ would become $M=\begin{pmatrix}-1&-1&-1&-1&-1\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\1&1&1&1&1\end{pmatrix}$,$\epsilon_1=\begin{pmatrix}-1\\1\\0\\0\\0\end{pmatrix}$, and then respectively for $\epsilon_2,\epsilon_3, ...$ or should I solve with $V=\begin{pmatrix}a\\b\\c\\d\\e\end{pmatrix}$ ? Because I don't understand what they're looking for if it's a general answer or a simple base to create.

because what I could find was my new basis (direcltly made into a base changing matrix because it's hard to type everything) : $P=\begin{pmatrix}-1&-1&-1&-1&-1\\1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\end{pmatrix}$

then I solved and answered $M'=P^-1MP=\begin{pmatrix}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&-1\\0&0&0&0&0\end{pmatrix}$

Then, right after this question, the big problem:

Question: now, $V\notin Ker(f)$ I have to find a basis so B matrix would be : B = \begin{pmatrix}\gamma&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{pmatrix}

and of course I need to precise $\gamma$

So here, I thought I could use $V=\begin{pmatrix}-1\\0\\0\\0\\0\end{pmatrix}$ and keep my $Ker(f)$ basis I found before and use the same formula $B=P^-1M'P$ but it didn't sound good at all ... Because my $\gamma=0$ which is nonsense ... I'm a bit confused and it makes me stuck

Thank you in advance and sorry for the big stuff.

Brocolus
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1 Answers1

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Hints:

Question 1

I would assume this question requires a general answer, so setting a value for $V$ is incorrect.

Before you can find the matrix of $f$ in the new basis, you need to find $\epsilon_5$. Since $\mathbb{R}^5$ has dimension 5, it is enough that $\{\epsilon_1, \epsilon_2, \epsilon_3, \epsilon_4, \epsilon_5\}$ be linearly independent. How can we choose $\epsilon_5$ to make this true? Remember $\{\epsilon_1, \epsilon_2, \epsilon_3, \epsilon_4\}$ spans the nullspace of $f$. Let's take $\epsilon_5$ to be some arbitrary vector not in $\mathtt{Ker}(f)$. Then, if $$\lambda_1\epsilon_1 + \lambda_2\epsilon_2 + \lambda_3\epsilon_3 + \lambda_4\epsilon_4 + \lambda_5\epsilon_5 = 0,$$ then $$f(\lambda_1\epsilon_1 + \lambda_2\epsilon_2 + \lambda_3\epsilon_3 + \lambda_4\epsilon_4 + \lambda_5\epsilon_5) = \lambda_1f(\epsilon_1) + \lambda_2f(\epsilon_2) + \lambda_3f(\epsilon_3) + \lambda_4f(\epsilon_4) + \lambda_5f(\epsilon_5) = 0$$ $$\lambda_1f(\epsilon_5) = 0.$$

Since $\epsilon_5 \not\in \mathtt{Ker}(f)$, we have that $\lambda_5 = 0$. Thus,

$$\lambda_1\epsilon_1 + \lambda_2\epsilon_2 + \lambda_3\epsilon_3 + \lambda_4\epsilon_4 = 0.$$

But the only way for this to happen is if $\lambda_1 = \lambda_2 = \lambda_3 = \lambda_4 = 0$ (why?) so $\{\epsilon_1, \epsilon_2, \epsilon_3, \epsilon_4, \epsilon_5\}$ is linearly independent.

Now, remember that the matrix of $f$ with respect to this basis will be $\left[f(\epsilon_1)|f(\epsilon_2)|f(\epsilon_3)|f(\epsilon_4)|f(\epsilon_5)\right]$.

Question 2

We observe that $f(V) = MV = \left[\begin{array}{ccccc} a & a & a & a & a \\ b & b & b & b & b \\ c & c & c & c & c \\ d & d & d & d & d \\ e & e & e & e & e \\ \end{array}\right] \left[\begin{array}{c} a \\ b \\ c \\ d \\ e \\ \end{array}\right] = a\left[\begin{array}{c} a \\ b \\ c \\ d \\ e \\ \end{array}\right] + b\left[\begin{array}{c} a \\ b \\ c \\ d \\ e \\ \end{array}\right] + c\left[\begin{array}{c} a \\ b \\ c \\ d \\ e \\ \end{array}\right] + d\left[\begin{array}{c} a \\ b \\ c \\ d \\ e \\ \end{array}\right] + e\left[\begin{array}{c} a \\ b \\ c \\ d \\ e \\ \end{array}\right] = (a+b+c+d+e)\left[\begin{array}{c} a \\ b \\ c \\ d \\ e \\ \end{array}\right]$

What does this tell use about the basis we need and the value of $\gamma$?

  • Thank you for your help !

    Question 1 : So I tried the vector $\epsilon_5 = \begin{pmatrix}-1\0\0\0\0\end{pmatrix}$ and it seemed to be working when I want to verify the family is linearly independant $\lambda_1\epsilon_1+...+\lambda_5\epsilon_5 = 0 \implies \lambda_1 = \lambda_2 = ... = \lambda_5 = 0$ and $\epsilon_5 \notin Ker(f)$ so i would say we need to have an $\epsilon_5$ vector with $a\neq 0$ and $b=c=d=e=0$ ?

    Question 2 : alright i tried $MV = BV \iff \gamma * a = a*(a+b+c+d+e) \iff \gamma = a+b+c+d+e$ because $a\neq 0$

     – Brocolus Jun 22 '14 at 23:48
  • sorry I just think I wrote some crap... so $MV=a(a+b+c+d+e)$ and $BV=\gamma a$ but then I don't know how can I relate them together – Brocolus Jun 23 '14 at 00:02
  • I've edited my answer to include a little more detail; tell me if it helps. I would say that for question 1, you don't need to be quite as specific as you are in choosing $\epsilon_5$ (although there is nothing wrong with being specific.) –  Jun 23 '14 at 04:20
  • I think I get it better, let's see : Question 1: okay $\lambda_5 = 0$ because the other $\epsilon$ family is in $Ker(f)$ when we apply $f$. Then $\lambda_1 = \lambda_2 = \lambda_3 = \lambda_4 =0$ because the family ${\epsilon_1 , \epsilon_2 , \epsilon_3 , \epsilon_4}$ is known as a basis with $V=\epsilon_4$ and the other epsilon aren't equal to the null vector so it implies this family is linearly independant. Question 2 : so what I notice is that the basis I would use to get $\gamma$ is $Vect(\begin{pmatrix}1\0\0\0\0\end{pmatrix},\epsilon_1,...,\epsilon_4)$ and $\gamma = a$ ? – Brocolus Jun 23 '14 at 09:45
  • So it means my B matrix basis would be $Vect(f(\begin{pmatrix}1\0\0\0\0\end{pmatrix}),f(\epsilon_1),...,f(\epsilon_4))$ which looks like what we are looking for – Brocolus Jun 23 '14 at 09:52
  • You're correct on question 1. For question 2, remember that the matrix $B$ is with respect to some new basis.
    $$f\left(\left[\begin{array}{c}1\0\0\0\0\end{array}\right]\right) =\left[ \begin{array}{c}a\a\a\a\a\end{array}\right] \not= \gamma\left[\begin{array}{c}1\0\0\0\0\end{array}\right]$$ for any $\gamma$, so $$\left[\begin{array}{c}1\0\0\0\0\end{array}\right]$$ cannot be the first vector of the basis. We need a vector $u$ such that $$f(u) = \gamma u$$ for some $\gamma$.     –  Jun 23 '14 at 20:24
  • hmm I when I calculate $f\begin{pmatrix}1\0\0\0\0\end{pmatrix}$ : $\begin{pmatrix} a&a&a&a&a\ b&b&b&b&b\ c&c&c&c&c\ d&d&d&d&d\ e&e&e&e&e\ \end{pmatrix}\begin{pmatrix}1\0\0\0\0\end{pmatrix}=\begin{pmatrix}a\b\c\d\e\end{pmatrix}$ but otherwise, I'll try this : let $X=\begin{pmatrix}x\y\z\t\u\end{pmatrix}$, $MX=\gamma X $ $\Longleftrightarrow$ $\begin{cases}a(x+y+z+t+u)=\gamma x\b(x+y+z+t+u)=\gamma y\c(x+y+z+t+u)=\gamma z\d(x+y+z+t+u)=\gamma t\e(x+y+z+t+u)=\gamma u \end{cases}$ but it's impossible to solve ! I really have no idea ... – Brocolus Jun 23 '14 at 22:25
  • Personnally I found : $\begin{pmatrix} a&a&a&a&a\ b&b&b&b&b\ c&c&c&c&c\ d&d&d&d&d\ e&e&e&e&e\ \end{pmatrix} \begin{pmatrix}1\0\0\0\0\end{pmatrix}=\begin{pmatrix}a\b\c\d\e\end{pmatrix}$ – Brocolus Jun 23 '14 at 22:31
  • So yea anyway it doesn't work but I really have no idea how it can work because I guess I should prove that $Det(M-\lambda I_5) = (\lambda - \gamma)\lambda ^2$ – Brocolus Jun 23 '14 at 22:38
  • I miscalculated in the last comment: I should have $$f\left(\left[\begin{array}{c}1\0\0\0\0\end{array}\right]\right) = \left[\begin{array}{c}a\b\c\d\e\end{array}\right]$$ as you have written. To find the needed vector, calculate $f(V)$, as I have done in the answer, and think about what that result tells us. –  Jun 24 '14 at 02:04
  • okay so if I calculate $f(V)=\gamma V$ so $MV= \begin{pmatrix}a&a&a&a&a&a\b&b&b&b&b\c&c&c&c&c\d&d&d&d&d\e&e&e&e&e\end{pmatrix} \begin{pmatrix}a\b\c\d\e\end{pmatrix}$ = $\begin{cases}a(a+b+ c+d+e)=\gamma a\b(a+b+c+d+e)=\gamma b\c(a+b+c+d+e)=\gamma c\d(a+b+c+d+e)=\gamma d\e(a+b+c+d+e)=\gamma e\end{cases}$ $\iff$ $\gamma = a+b+c+d+e$ – Brocolus Jun 24 '14 at 18:35
  • oh wait so it should be : $\gamma = a+b+c+d+e$ and the basis : $Vect(f\begin{pmatrix}a\b\c\d\e\end{pmatrix},\epsilon_1,\epsilon_2,\epsilon_3, \epsilon_4)$ – Brocolus Jun 25 '14 at 09:24
  • That works, yes. You could also do ${V, \epsilon_1, \epsilon_2, \epsilon_3, \epsilon_4}$ for the basis. –  Jun 26 '14 at 01:42
  • whooooo that was a long way but it worked ! Thank you very much for your help ! – Brocolus Jun 26 '14 at 13:21