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[SOLVED] matrices......whose null space consists all linear combinations

karush

Well-known member
Jan 31, 2012
3,066
$
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]
$
 
Last edited:

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
821
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.
 

karush

Well-known member
Jan 31, 2012
3,066
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:
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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. :oops:

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.