How to determine the bases B and C

[mathmari]

Hey!

We have tha matrix \begin{equation*}A:=\begin{pmatrix}1 & 5 & 8 & 2 \\ 2 & 4 & 6 & 0 \\ 3 & 3 & 8 & 2 \\ 4 & 2 & 6 & 0 \\ 5 & 1 & 8 & 2\end{pmatrix}\in \mathbb{R}^{5\times 4}\end{equation*}

Determine a basis $B$ of $\mathbb{R}^4$ and a basis $C$ of $\mathbb{R}^5$ such that $M_C^B(A)$ contains at the left upper corner the unit matrix and everywhere else zeroes. Could you give me a hint for that? :unsure:

Do we have to write each column of $A$ as a linear combination of $C$ and the corresponding coefficients have to satisfy the desired form that $M$ will have? :unsure:

 

[I like Serena]

Hey mathmari!

How about Singular Value Decomposition? (Wondering)

We can write $A=U\Sigma V^T$, where $U$ and $V$ are orthogonal matrices, and $\Sigma$ is a diagonal matrix.
Furthermore $\Sigma$ has first positive entries on its diagonal, followed by zero entries.
Suppose there are $k$ positive entries,
We can then write $\Sigma = DI_k D$, where $D$ is a diagonal matrix, and $I_k$ is a zero matrix with an identity matrix at the left top of size $k$.

The basises $B$ a $C$ correspond to $UD$ and $VD$, and $I_k$ is the desired $M_C^B(A)$ matrix.

 

[mathmari]

We can write $A=U\Sigma V^T$, where $U$ and $V$ are orthogonal matrices, and $\Sigma$ is a diagonal matrix.
Furthermore $\Sigma$ has first positive entries on its diagonal, followed by zero entries.
Suppose there are $k$ positive entries,
We can then write $\Sigma = DI_k D$, where $D$ is a diagonal matrix, and $I_k$ is a zero matrix with an identity matrix at the left top of size $k$.

The basises $B$ a $C$ correspond to $UD$ and $VD$, and $I_k$ is the desired $M_C^B(A)$ matrix.

Could you explain that method further to me? I haven't really understood that. :unsure:

 

[I like Serena]

Could you explain that method further to me? I haven't really understood that.

Let $\mathbf b_1$ be the first vector of the basis $B$, which is represented by standard vector $\mathbf e_1$ with respect to $B$.
Let $\mathbf c_1$ be the first vector of the basis $C$, which is represented by standard vector $\mathbf f_1$ with respect to $C$.

Suppose that the top left entry of $M_C^B(A)$ is $1$.
Then $M_C^B(A) \mathbf e_1 = \mathbf f_1$ and $A\mathbf b_1 = \mathbf c_1$ yes?
Furthermore $B\mathbf e_1 = \mathbf b_1$ and $C\mathbf f_1 = \mathbf c_1$ so that $AB\mathbf e_1 = C\mathbf f_1$.

Let $(d_i)$ be the vector that represents the diagonal of $M_C^B(A)$.
That is, it starts with $1$'s and ends with $0$'s.

Now repeat for each basis vector $\mathbf b_i$ in $B$ ($i=1,\ldots,4$).
Then we get $M_C^B(A) \mathbf e_i = d_i\mathbf f_i$ and $A \mathbf b_i = d_i\mathbf c_i$.
Furthermore $B\mathbf e_i = \mathbf b_i$ ($i=1,\ldots,4$) and $C\mathbf f_j = \mathbf c_j$ ($j=1,\ldots,5$).

Rewrite as $AB\mathbf e_i = d_i C\mathbf f_i \implies AB\mathbf e_i = C \mathbf f_i d_i \implies ABI_4 = CI_5M_C^B(A) \implies AB = CM_C^B(A) \implies A = CM_C^B(A)B^{-1}$.
Right? (Sweating)

 

[mathmari]

Let $\mathbf b_1$ be the first vector of the basis $B$, which is represented by standard vector $\mathbf e_1$ with respect to $B$.
Let $\mathbf c_1$ be the first vector of the basis $C$, which is represented by standard vector $\mathbf f_1$ with respect to $C$.

Suppose that the top left entry of $M_C^B(A)$ is $1$.
Then $M_C^B(A) \mathbf e_1 = \mathbf f_1$ and $A\mathbf b_1 = \mathbf c_1$ yes?
Furthermore $B\mathbf e_1 = \mathbf b_1$ and $C\mathbf f_1 = \mathbf c_1$ so that $AB\mathbf e_1 = C\mathbf f_1$.

Let $(d_i)$ be the vector that represents the diagonal of $M_C^B(A)$.
That is, it starts with $1$'s and ends with $0$'s.

Now repeat for each basis vector $\mathbf b_i$ in $B$ ($i=1,\ldots,4$).
Then we get $M_C^B(A) \mathbf e_i = d_i\mathbf f_i$ and $A \mathbf b_i = d_i\mathbf c_i$.
Furthermore $B\mathbf e_i = \mathbf b_i$ ($i=1,\ldots,4$) and $C\mathbf f_j = \mathbf c_j$ ($j=1,\ldots,5$).

Rewrite as $AB\mathbf e_i = d_i C\mathbf f_i \implies AB\mathbf e_i = C \mathbf f_i d_i \implies ABI_4 = CI_5M_C^B(A) \implies AB = CM_C^B(A) \implies A = CM_C^B(A)B^{-1}$.
Right? (Sweating)

Can we not just take for example as $B$ the standardbasis of $\mathbb{R}^4$ and then from $A = CM_C^B(A)B^{-1}$ to solve for $C$ ? Or do we have to apply the method you mentioned in post #2 ? :unsure:

 

[I like Serena]

Can we not just take for example as $B$ the standardbasis of $\mathbb{R}^4$ and then from $A = CM_C^B(A)B^{-1}$ to solve for $C$ ? Or do we have to apply the method you mentioned in post #2 ?

We can indeed do it more or less like that.

However, we have to distinguish vectors in the null space from vectors in the row space.
That is, we can choose $B$ to consist of a couple of arbitrary independent (standard unit) vectors that span the row space of A, followed by a basis of the null space of A.
Or alternatively we can choose $C$ to consist of arbitrary independent (standard unit) vectors that span the row space of $A^T$, followed by a basis of the null space of $A^T$.

Suppose the rank of $A$ is $2$.
Then suppose $\{b_1,b_2\}$ is a basis of the row space of $A$, suppose $\{b_3,b_4\}$ is a basis of the null space of $A$, and suppose $\{c_3,c_4,c_5\}$ is a basis of the null space of $A^T$.
We must pick $c_1=Ab_1$ and $c_2=Ab_2$.
Then we can verify that $B$ and $C$ satisfy the condition by verifying that $CM_C^B(A)=AB$.

Either way, we will have to find the null space of $A$ and the null space of $A^T$. Can we find them? (Sweating)

 

[HOI]

Hey!

We have tha matrix \begin{equation*}A:=\begin{pmatrix}1 & 5 & 8 & 2 \\ 2 & 4 & 6 & 0 \\ 3 & 3 & 8 & 2 \\ 4 & 2 & 6 & 0 \\ 5 & 1 & 8 & 2\end{pmatrix}\in \mathbb{R}^{5\times 4}\end{equation*}

Determine a basis $B$ of $\mathbb{R}^4$ and a basis $C$ of $\mathbb{R}^5$ such that $M_C^B(A)$ contains at the left upper corner the unit matrix and everywhere else zeroes.

Do you understand what this MEANS? IF there exist such bases then the basis of $R
4$ must be such that A maps some of the basis vectors to themselves (how many depending on the size of the "unit matrix at the left upper corner") and maps the other basis vectors to 0.

You might start by solving Ax= x or by solving Ax= 0.2
.'0?632

 

[HOI]

Okay, I did a row reduction and reduced the matrix to
$\begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$.

So the upper left corner is the 3 by 3 identity matrix. There exist a 3 dimensional subspace of $R^4$ on which Ax= x and Ax= 0 on the orthogonal complement.

 

[I like Serena]

I did row reduction as well and found $\begin{pmatrix}1&5&8&2\\&3&5&2\\&&1&1\\{}\\{}\end{pmatrix}$.
It follows that the null space of $A$ is the span of $\begin{pmatrix}1\\1\\-1\\1\end{pmatrix}$.

It means that we can pick $b_1,b_2,b_3$ arbitrary, and that we must have that $b_4=\begin{pmatrix}1\\1\\-1\\1\end{pmatrix}$ or a non-zero multiple.

I also did row reduction on $A^T$ and found $\begin{pmatrix}1&2&3&4&5\\&1&2&3&4\\&&1&1&2\\{}\end{pmatrix}$.
It follows that the null space of $A^T$ is the span of $\begin{pmatrix}-1\\2\\0\\-2\\1\end{pmatrix}$ and $\begin{pmatrix}1\\-1\\-1\\1\\0\end{pmatrix}$.

If means that we can pick $c_1,c_2,c_3$ arbitrary with the restriction that $Ab_i=c_i$ for $i=1,2,3$, and we can pick $c_4=\begin{pmatrix}-1\\2\\0\\-2\\1\end{pmatrix}$ and $c_5=\begin{pmatrix}1\\-1\\-1\\1\\0\end{pmatrix}$.