# Matrices.......whose null space consists all linear combinations

• MHB
• karush
In summary: It just has to have $x$ in the null space.So any matrix with $4$ independent rows will do.Even a $1000$ by $5$ matrix. :rofl:wow that's a lot of choicesIn summary, the task is to construct 7 matrices not yet in row reduced echelon form whose null space consists of all linear combinations of either just x, just y, or just z, or pairs of x and y, y and z, or z and x, or all three vectors x, y, and z.
karush
Gold Member
MHB
$v=\left[\begin{array}{r} -3\\-4\\-5\\4\\-1 \end{array}\right] w=\left[\begin{array}{r} -2\\0 \\1 \\4 \\-1 \end{array}\right] x=\left[\begin{array}{r} 2\\3 \\4 \\-5 \\0 \end{array}\right] y=\left[\begin{array}{r} -2\\1 \\0 \\-2 \\7 \end{array}\right] z=\left[\begin{array}{r} -1\\0 \\2 \\-3 \\5 \end{array}\right]$
Construct matrices not yet row reduced echelon form whose null space consists all linear combinations of
1. just x
2. just y
3. just z
ok I presume this

$A_1=a_1\left[\begin{array}{r}2\\3 \\4 \\-5 \\0\end{array}\right] =\left[\begin{array}{r}2a_1\\3a_1 \\4a_1 \\-5a_1 \\0\end{array}\right]$

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Okay, so you want to take $a_1= 0$ so that 0 times $\begin{bmatrix}2 \\ 3 \\ 4 \\ -5 \\ 0\end{bmatrix}$ is 0. That is a valid answer to that question. But I suspect that is not what they really meant! I suspect they want a square, 5 by 5, matrix. In that case you want
$\begin{bmatrix} a_1 & a_2 & a_3 & a_4 & a_5 \\ b_1 & b_2 & b_3 & b_4 & b_5 \\ c_1 & c_2 & c_3 & c_4 & c_5 \\ d_1 & d_2 & d_3 & d_4 & d_5 \\ e_1 & e_2 & e_3 & e_4 & e_5 \end{bmatrix} $$\begin{bmatrix}2 \\ 3 \\ 4 \\ -5 \\ 0\end{bmatrix}=$$\begin{bmatrix}2a_1+ 3a_2+ 4a_4- 5a_5 \\ 2b_1+ 3b_2+ 4b_4- 5b_5 \\ 2c_1+ 3c_2+ 4c_4- 5c_5 \\ 2d_1+ 3d_2+ 4d_4- 5d_5 \\ 2e_1+ 3e_2+ 4e_4- 5e_5 \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$.

That is five equations in 25 unknowns so there is no unique solution. You can choose 20 of them to be any numbers you want (I would probably use 0's and 1's) and solve for the remaining 5.

ok I think that would answer v,w,x,y,z
altho I didn't mention it in the OP i thot if just one matrix was ok the rest of the combinations would be just additional modifications

Construct 7 matrices not yet row reduced echelon form whose null space consists all linear combinations of
1. just x
2. just y
3. just z
4. of the pairs x and y
5. of the pairs y and z
6. of the pairs z and x
7. all 3 vectors x y and z

So the pairs of x an y would be
$A_1=a_1\left[\begin{array}{r}2\\3 \\4 \\-5 \\0\end{array}\right] + a_2\left[\begin{array}{r} -2\\1 \\0 \\-2 \\7 \end{array}\right] =\left[\begin{array}{rl} 2a_1&+(-2a_2)\\ 3a_1&+ a_2 \\ 4a_1&+(-2a_2) \\ -5a_1&+7a_2 \end{array}\right]$
and so on ,,, hopefully:unsure:

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Country Boy said:
Okay, so you want to take $a_1= 0$ so that 0 times $\begin{bmatrix}2 \\ 3 \\ 4 \\ -5 \\ 0\end{bmatrix}$ is 0. That is a valid answer to that question.
Not quite. They are asking for a matrix such that the null space is "just" multiples of $x$.
If we take a null matrix, then vectors independent to $x$ are also in the null space, which violates the condition.

Instead we need a matrix $A_1$ of rank $4$ such that $A_1x=0$.
At a minimum the matrix must have $4$ independent rows.
Furthermore, each row must be perpendicular to $x$.
If we want to, we can add more rows, which must then be linear combinations of the $4$ independent rows.

So let's pick $4$ rows that are independent and perpendicular to $x$.
For instance:
$$A_1 x = \begin{bmatrix}5&0&0&2&0 \\ 0&5&0&3&0 \\ 0&0&5&4&0 \\ 0&0&0&0&1\end{bmatrix}\begin{bmatrix}2 \\ 3 \\ 4 \\ -5 \\ 0\end{bmatrix}$$
We can verify that each of the $5$ unit vectors are indeed not in the null space.
From the row echelon form we can see that the rows are indeed independent so that its rank is $4$ as needed.
And since each row is perpendicular to $x$, we have that $x$ is in the null space.

Oh, and oops, it's already in row echelon form although that was not needed.

Country Boy said:
But I suspect that is not what they really meant! I suspect they want a square, 5 by 5, matrix.
So no, it doesn't have to be a 5x5 matrix.

## 1. What is a matrix?

A matrix is a rectangular array of numbers or symbols arranged in rows and columns. It is commonly used in mathematics, physics, and engineering to represent and manipulate data.

## 2. What is the null space of a matrix?

The null space of a matrix is the set of all vectors that when multiplied by the matrix result in the zero vector. In other words, it is the set of all solutions to the equation Ax = 0, where A is the given matrix.

## 3. What is a linear combination?

A linear combination is a mathematical operation where two or more vectors are multiplied by constants and then added together. For example, given vectors u and v and constants a and b, the linear combination au + bv is a new vector that is a combination of u and v.

## 4. How do you determine if a matrix's null space consists of all linear combinations?

To determine if a matrix's null space consists of all linear combinations, you can perform row reduction on the matrix and check if the resulting matrix has a row of zeros. If it does, then the null space consists of all linear combinations of the remaining rows.

## 5. Why is the null space of a matrix important?

The null space of a matrix is important because it provides information about the solutions to a system of linear equations represented by the matrix. It can also be used to determine the rank of a matrix and to find a basis for the column space of the matrix.

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