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M has the unit matrix at the upper left side and zero everywhere else

mathmari

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Apr 14, 2013
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Hey!! :giggle:

Let $\lambda\in \mathbb{R}$ and \begin{equation*}a=\begin{pmatrix}1 & 2 &-1& \lambda & -\lambda \\ 0 & 1 & -1& \lambda & 2\\ 2 & 2 & 1 & 1 & 3\lambda-1 \\ 1 & 1 & 1 & \lambda & 5\end{pmatrix}\in \mathbb{R}^{4\times 5}\end{equation*}
(a) Let $\lambda=1$. Determine a Basis $\mathcal{B}$ of $\mathbb{R}^5$ and a Basis $\mathcal{C}$ of $\mathbb{R}^4$, such that $\mathcal{M}_{\mathcal{C}}^{\mathcal{B}}(\phi_a)$ at the upper left side there is an unit matrix and elsewise zero.

(b) Determine the rank of $a$ and $a^T$.



For question (a) :

With $\lambda=1$ we get the matrix \begin{equation*}a=\begin{pmatrix}1 & 2 &-1& 1 & -1 \\ 0 & 1 & -1& 1 & 2\\ 2 & 2 & 1 & 1 & 2 \\ 1 & 1 & 1 & 1 & 5\end{pmatrix}\end{equation*}
Let $\mathcal{B}=\{b_1 , b_2, b_3, b_4, b_5\}$ and $\mathcal{C}=\{c_1 , c_2, c_3, c_4\}$.

It holds that \begin{equation*}\mathcal{M}_{\mathcal{C}}^{\mathcal{B}}(\phi_a)=\left (\gamma_{\mathcal{C}}(\phi_a(b_1))\mid \gamma_{\mathcal{C}}(\phi_a(b_2))\mid \gamma_{\mathcal{C}}(\phi_a(b_3)) \mid \gamma_{\mathcal{C}}(\phi_a(b_4)) \mid \gamma_{\mathcal{C}}(\phi_a(b_5))\right )\end{equation*}
It holds also that $\gamma_{\mathcal{C}}(\phi_a(b_i))=\begin{pmatrix}\alpha_1 \\ \alpha_2\\ \alpha_3\\ \alpha_4\end{pmatrix}$ with $\phi_a(b_i)=\alpha_1\cdot c_1+\alpha_2\cdot c_2+\alpha_3\cdot c_3+\alpha_4\cdot c_4$.

That $\mathcal{M}_{\mathcal{C}}^{\mathcal{B}}(\phi_a)$ at the upper left side there is an unit matrix and elsewise zero, does it mean it must be of the form \begin{equation*}a=\begin{pmatrix}1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0& 0 & 0\\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0\end{pmatrix}\end{equation*} ? Or what does this mean?


:unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
That $\mathcal{M}_{\mathcal{C}}^{\mathcal{B}}(\phi_a)$ at the upper left side there is an unit matrix and elsewise zero, does it mean it must be of the form \begin{equation*}a=\begin{pmatrix}1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0& 0 & 0\\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0\end{pmatrix}\end{equation*} ? Or what does this mean?
Hey mathmari !!

That's how I interpret it as well. (Nod)
 

mathmari

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Apr 14, 2013
4,713

Klaas van Aarsen

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Mar 5, 2012
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But how can we find the desired basis? Could you give me a hint?
We must have for $i=1,\ldots,4$ that $a b_i = c_i$, and additionally that $a b_5=0$.
Furthermore, the $b_i$ must be independent, and the $c_i$ must be independent as well.
Beyond that we are free to pick $b_i$ and $c_i$ however we want.

Suppose we pick the $b_i$ to be unit vectors?
What will we get for the $c_i$ then? 🤔
 
Last edited:

mathmari

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Apr 14, 2013
4,713
We must have for $i=1,\ldots,4$ that $a b_i = c_i$, and additionally that $a b_5=0$.
Furthermore, the $b_i$ must be independent, and the $c_i$ must be independent as well.
Beyond that we are free to pick $b_i$ and $c_i$ however we want.

Suppose we pick the $b_i$ to be unit vectors?
What will we get for the $c_i$ then? 🤔
When we pick the $b_i$ (you mean for $i=1,2,3,4$, or not?) to be unit vectors we have \begin{equation*}c_1=ab_1=\begin{pmatrix}1 \\ 0 \\ 2 \\ 1\end{pmatrix}, \ c_2=ab_2=\begin{pmatrix}2 \\ 1 \\ 2 \\ 1\end{pmatrix}, \ c_3=ab_3=\begin{pmatrix}-1 \\ -1 \\ 1 \\ 1\end{pmatrix}, \ c_4=ab_4=\begin{pmatrix}1 \\ 1 \\ 1 \\ 1\end{pmatrix}\end{equation*} So we have to check if these are linearly independent, if not we have to use the vectors $b_i$, right?

The vector $b_5$ must be such that $a b_5=0$ and independent from $b_1, \ldots , b_4$ so it cannot be a unit vector, can it?

:unsure:
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
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So we have to check if these are linearly independent, if not we have to use the vectors $b_i$, right?

The vector $b_5$ must be such that $a b_5=0$ and independent from $b_1, \ldots , b_4$ so it cannot be a unit vector, can it?
I guess we will have to find a non-zero vector $b_5$ that maps to the null vector. 🤔
As long as the other 4 vectors are independent from each other and from $b_5$ they should be good to go. 🤔
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,713
I guess we will have to find a non-zero vector $b_5$ that maps to the null vector. 🤔
As long as the other 4 vectors are independent from each other and from $b_5$ they should be good to go. 🤔
\begin{align*}\begin{pmatrix}1 & 2 &-1& 1 & -1 \\ 0 & 1 & -1& 1 & 2\\ 2 & 2 & 1 & 1 & 2 \\ 1 & 1 & 1 & 1 & 5\end{pmatrix}& \ \overset{R_4:R_4-R_1}{\longrightarrow } \ \begin{pmatrix}1 & 2 &-1& 1 & -1 \\ 0 & 1 & -1& 1 & 2\\ 2 & 2 & 1 & 1 & 2 \\ 0 & -1 & -2 & 0 & 6\end{pmatrix}\\ & \ \overset{R_3:R_3-2\cdot R_1}{\longrightarrow } \ \begin{pmatrix}1 & 2 &-1& 1 & -1 \\ 0 & 1 & -1& 1 & 2\\ 0 & -2 & 3 & -1 & 4 \\ 0 & -1 & -2 & 0 & 6\end{pmatrix} \\ & \ \overset{R_4:R_4+R_2}{\longrightarrow } \ \begin{pmatrix}1 & 2 &-1& 1 & -1 \\ 0 & 1 & -1& 1 & 2\\ 0 & -2 & 3 & -1 & 4 \\ 0 & 0 & -3 & 1 & 8\end{pmatrix} \\ & \ \overset{R_3:R_3+2\cdot R_2}{\longrightarrow } \ \begin{pmatrix}1 & 2 &-1& 1 & -1 \\ 0 & 1 & -1& 1 & 2\\ 0 & 0 & 1 & 1 & 8 \\ 0 & 0 & -3 & 1 & 8\end{pmatrix}\\ & \ \overset{R_4:R_4+3\cdot R_3}{\longrightarrow } \ \begin{pmatrix}1 & 2 &-1& 1 & -1 \\ 0 & 1 & -1& 1 & 2\\ 0 & 0 & 1 & 1 & 8 \\ 0 & 0 & 0 & 4 & 32\end{pmatrix}\end{align*}
We get the equations \begin{align*}a+2b-c+d-e=&0 \\ b-c+d+2e =&0 \\ c+d+8e=&0 \\ 4d+32e=&0\end{align*}
Choosing for eample $e=1$ we get the vector $b_5=\begin{pmatrix}-3 \\ 6 \\ 0 \\ -8 \\ 1\end{pmatrix}$.

So the basis $B$ is $\left \{\begin{pmatrix}1\\ 0 \\ 0\\ 0\\ 0\end{pmatrix}, \ \begin{pmatrix}0\\ 1 \\ 0\\ 0\\ 0\end{pmatrix}, \ \begin{pmatrix}0\\ 0 \\ 1\\ 0\\ 0\end{pmatrix} , \ \begin{pmatrix}0\\ 0 \\ 0\\ 1\\ 0\end{pmatrix}, \ \begin{pmatrix}-3 \\ 6 \\ 0 \\ -8 \\ 1\end{pmatrix}\right \}$.

And the basis $C$ is $\left \{\begin{pmatrix}1 \\ 0 \\ 2 \\ 1\end{pmatrix}, \ \begin{pmatrix}2 \\ 1 \\ 2 \\ 1\end{pmatrix}, \ \begin{pmatrix}-1 \\ -1 \\ 1 \\ 1\end{pmatrix}, \ \begin{pmatrix}1 \\ 1 \\ 1 \\ 1\end{pmatrix}\right \}$, i.e. the columns of the matrix $a$, which are linearly independent as at the row echelon form there is no zero-row.

Is that correct? :unsure:
 

Klaas van Aarsen

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Mar 5, 2012
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Looks correct to me. (Nod)
 

mathmari

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Apr 14, 2013
4,713