Mixing
Determining the rank, nullspace and image of a matrix with one unknown
$begingroup$
Consider the matrix:
$$A=begin{pmatrix}x &0&2&2 \-2&2&-2&2\2&2&-2&2\2&0&2&-2 end{pmatrix}$$
Determine $text{rank(A)}$ as well as a basis for the nullspace and image of $A$.
My attempt:
$$text{ref}(A)=begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&0&0&-4 end{pmatrix}$$
Since $text{rank}(A)=text{rank}(text{ref}(A)) implies text{rank}(A)=4$
$$implies text{dim}(text{im(A)})=4 \ implies text{dim}(text{null(A)})=0$$
Therefore a basis for the image could just be my column vectors of $A$ and a basis for the nullspace would be $(0,0,0,0)^T$. Would that be correct? Also, it seems like the rank, nullspace and image are totally independent of $x$. This is a bit puzzling since the question on my problem set specifically askas how they depend on $x$. Am I doing something wrong?
Edit 1: Here are the row operations I performed:
$$begin{pmatrix}x &0&2&2 \-2&2&-2&2\2&2&-2&2\2&0&2&-2 end{pmatrix}xrightarrow{R_2 leftrightarrow R_1} begin{pmatrix}-2 &2&-2&2 \x&0&2&2\2&2&-2&2\2&0&2&-2 end{pmatrix}xrightarrow{R_2-frac{x}{2}R_3} begin{pmatrix}-2 &2&-2&2 \0&-x&2+x&2-x\2&2&-2&2\2&0&2&-2 end{pmatrix} \ xrightarrow{R_3+R1, R_4+R_1} begin{pmatrix}-2 &2&-2&2 \0&-x&2+x&2-x\0&4&-4&4\0&2&0&0 end{pmatrix} xrightarrow{R_3 leftrightarrow R_2} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&-x&2+x&2-x\0&2&0&0 end{pmatrix} \ xrightarrow{R_3+frac{x}{4}R_2} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&2&0&0 end{pmatrix} xrightarrow{R_4-frac{1}{2}R_2} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&&2&-2 end{pmatrix} xrightarrow{R_4-R_3} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&0&0&-4 end{pmatrix}$$
linear-algebra
asked Jan 20 at 12:27
NullspaceNullspace
374110
$endgroup$
|
show 8 more comments
$begingroup$
Consider the matrix:
$$A=begin{pmatrix}x &0&2&2 \-2&2&-2&2\2&2&-2&2\2&0&2&-2 end{pmatrix}$$
Determine $text{rank(A)}$ as well as a basis for the nullspace and image of $A$.
My attempt:
$$text{ref}(A)=begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&0&0&-4 end{pmatrix}$$
Since $text{rank}(A)=text{rank}(text{ref}(A)) implies text{rank}(A)=4$
$$implies text{dim}(text{im(A)})=4 \ implies text{dim}(text{null(A)})=0$$
Therefore a basis for the image could just be my column vectors of $A$ and a basis for the nullspace would be $(0,0,0,0)^T$. Would that be correct? Also, it seems like the rank, nullspace and image are totally independent of $x$. This is a bit puzzling since the question on my problem set specifically askas how they depend on $x$. Am I doing something wrong?
Edit 1: Here are the row operations I performed:
$$begin{pmatrix}x &0&2&2 \-2&2&-2&2\2&2&-2&2\2&0&2&-2 end{pmatrix}xrightarrow{R_2 leftrightarrow R_1} begin{pmatrix}-2 &2&-2&2 \x&0&2&2\2&2&-2&2\2&0&2&-2 end{pmatrix}xrightarrow{R_2-frac{x}{2}R_3} begin{pmatrix}-2 &2&-2&2 \0&-x&2+x&2-x\2&2&-2&2\2&0&2&-2 end{pmatrix} \ xrightarrow{R_3+R1, R_4+R_1} begin{pmatrix}-2 &2&-2&2 \0&-x&2+x&2-x\0&4&-4&4\0&2&0&0 end{pmatrix} xrightarrow{R_3 leftrightarrow R_2} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&-x&2+x&2-x\0&2&0&0 end{pmatrix} \ xrightarrow{R_3+frac{x}{4}R_2} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&2&0&0 end{pmatrix} xrightarrow{R_4-frac{1}{2}R_2} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&&2&-2 end{pmatrix} xrightarrow{R_4-R_3} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&0&0&-4 end{pmatrix}$$
linear-algebra
asked Jan 20 at 12:27
NullspaceNullspace
374110
$endgroup$
$begingroup$
This isn't reduced row echelon form. If you simplify it, you will find that you end up getting the identity matrix as your reduced echelon form. The takeaway should be that the rank, nullspace, and image are independent of x in this case.
$endgroup$
– Jack Moody
Jan 20 at 12:42
1
$begingroup$
@JackMoody I meant to say row echelon form. But I am glad I got it somewhat right when saying it is independent of $x$. Thanks for your help
$endgroup$
– Nullspace
Jan 20 at 12:50
1
$begingroup$
I would say that all you have done is OK to me.
$endgroup$
– idriskameni
Jan 20 at 12:52
1
$begingroup$
Where did $x$ go in your echelon form?
$endgroup$
– Bernard
Jan 20 at 13:20
1
$begingroup$
You're welcome. It's a pleasure to help!
$endgroup$
– Bernard
Jan 20 at 14:29
|
show 8 more comments
$begingroup$
Consider the matrix:
$$A=begin{pmatrix}x &0&2&2 \-2&2&-2&2\2&2&-2&2\2&0&2&-2 end{pmatrix}$$
Determine $text{rank(A)}$ as well as a basis for the nullspace and image of $A$.
My attempt:
$$text{ref}(A)=begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&0&0&-4 end{pmatrix}$$
Since $text{rank}(A)=text{rank}(text{ref}(A)) implies text{rank}(A)=4$
$$implies text{dim}(text{im(A)})=4 \ implies text{dim}(text{null(A)})=0$$
Therefore a basis for the image could just be my column vectors of $A$ and a basis for the nullspace would be $(0,0,0,0)^T$. Would that be correct? Also, it seems like the rank, nullspace and image are totally independent of $x$. This is a bit puzzling since the question on my problem set specifically askas how they depend on $x$. Am I doing something wrong?
Edit 1: Here are the row operations I performed:
$$begin{pmatrix}x &0&2&2 \-2&2&-2&2\2&2&-2&2\2&0&2&-2 end{pmatrix}xrightarrow{R_2 leftrightarrow R_1} begin{pmatrix}-2 &2&-2&2 \x&0&2&2\2&2&-2&2\2&0&2&-2 end{pmatrix}xrightarrow{R_2-frac{x}{2}R_3} begin{pmatrix}-2 &2&-2&2 \0&-x&2+x&2-x\2&2&-2&2\2&0&2&-2 end{pmatrix} \ xrightarrow{R_3+R1, R_4+R_1} begin{pmatrix}-2 &2&-2&2 \0&-x&2+x&2-x\0&4&-4&4\0&2&0&0 end{pmatrix} xrightarrow{R_3 leftrightarrow R_2} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&-x&2+x&2-x\0&2&0&0 end{pmatrix} \ xrightarrow{R_3+frac{x}{4}R_2} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&2&0&0 end{pmatrix} xrightarrow{R_4-frac{1}{2}R_2} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&&2&-2 end{pmatrix} xrightarrow{R_4-R_3} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&0&0&-4 end{pmatrix}$$
linear-algebra
asked Jan 20 at 12:27
NullspaceNullspace
374110
$endgroup$
Consider the matrix:
$$A=begin{pmatrix}x &0&2&2 \-2&2&-2&2\2&2&-2&2\2&0&2&-2 end{pmatrix}$$
Determine $text{rank(A)}$ as well as a basis for the nullspace and image of $A$.
My attempt:
$$text{ref}(A)=begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&0&0&-4 end{pmatrix}$$
Since $text{rank}(A)=text{rank}(text{ref}(A)) implies text{rank}(A)=4$
$$implies text{dim}(text{im(A)})=4 \ implies text{dim}(text{null(A)})=0$$
Therefore a basis for the image could just be my column vectors of $A$ and a basis for the nullspace would be $(0,0,0,0)^T$. Would that be correct? Also, it seems like the rank, nullspace and image are totally independent of $x$. This is a bit puzzling since the question on my problem set specifically askas how they depend on $x$. Am I doing something wrong?
Edit 1: Here are the row operations I performed:
$$begin{pmatrix}x &0&2&2 \-2&2&-2&2\2&2&-2&2\2&0&2&-2 end{pmatrix}xrightarrow{R_2 leftrightarrow R_1} begin{pmatrix}-2 &2&-2&2 \x&0&2&2\2&2&-2&2\2&0&2&-2 end{pmatrix}xrightarrow{R_2-frac{x}{2}R_3} begin{pmatrix}-2 &2&-2&2 \0&-x&2+x&2-x\2&2&-2&2\2&0&2&-2 end{pmatrix} \ xrightarrow{R_3+R1, R_4+R_1} begin{pmatrix}-2 &2&-2&2 \0&-x&2+x&2-x\0&4&-4&4\0&2&0&0 end{pmatrix} xrightarrow{R_3 leftrightarrow R_2} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&-x&2+x&2-x\0&2&0&0 end{pmatrix} \ xrightarrow{R_3+frac{x}{4}R_2} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&2&0&0 end{pmatrix} xrightarrow{R_4-frac{1}{2}R_2} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&&2&-2 end{pmatrix} xrightarrow{R_4-R_3} begin{pmatrix}-2 &2&-2&2 \0&4&-4&4\0&0&2&2\0&0&0&-4 end{pmatrix}$$
linear-algebra
linear-algebra
asked Jan 20 at 12:27
NullspaceNullspace
374110
asked Jan 20 at 12:27
NullspaceNullspace
374110
edited Jan 20 at 13:54
Nullspace
asked Jan 20 at 12:27
NullspaceNullspace
374110
asked Jan 20 at 12:27
NullspaceNullspace
374110
asked Jan 20 at 12:27
NullspaceNullspace
374110
374110
$begingroup$
This isn't reduced row echelon form. If you simplify it, you will find that you end up getting the identity matrix as your reduced echelon form. The takeaway should be that the rank, nullspace, and image are independent of x in this case.
$endgroup$
– Jack Moody
Jan 20 at 12:42
1
$begingroup$
@JackMoody I meant to say row echelon form. But I am glad I got it somewhat right when saying it is independent of $x$. Thanks for your help
$endgroup$
– Nullspace
Jan 20 at 12:50
1
$begingroup$
I would say that all you have done is OK to me.
$endgroup$
– idriskameni
Jan 20 at 12:52
1
$begingroup$
Where did $x$ go in your echelon form?
$endgroup$
– Bernard
Jan 20 at 13:20
1
$begingroup$
You're welcome. It's a pleasure to help!
$endgroup$
– Bernard
Jan 20 at 14:29
|
show 8 more comments
$begingroup$
This isn't reduced row echelon form. If you simplify it, you will find that you end up getting the identity matrix as your reduced echelon form. The takeaway should be that the rank, nullspace, and image are independent of x in this case.
$endgroup$
– Jack Moody
Jan 20 at 12:42
1
$begingroup$
@JackMoody I meant to say row echelon form. But I am glad I got it somewhat right when saying it is independent of $x$. Thanks for your help
$endgroup$
– Nullspace
Jan 20 at 12:50
1
$begingroup$
I would say that all you have done is OK to me.
$endgroup$
– idriskameni
Jan 20 at 12:52
1
$begingroup$
Where did $x$ go in your echelon form?
$endgroup$
– Bernard
Jan 20 at 13:20
1
$begingroup$
You're welcome. It's a pleasure to help!
$endgroup$
– Bernard
Jan 20 at 14:29
$begingroup$
This isn't reduced row echelon form. If you simplify it, you will find that you end up getting the identity matrix as your reduced echelon form. The takeaway should be that the rank, nullspace, and image are independent of x in this case.
$endgroup$
– Jack Moody
Jan 20 at 12:42
$begingroup$
This isn't reduced row echelon form. If you simplify it, you will find that you end up getting the identity matrix as your reduced echelon form. The takeaway should be that the rank, nullspace, and image are independent of x in this case.
$endgroup$
– Jack Moody
Jan 20 at 12:42
1
1
$begingroup$
@JackMoody I meant to say row echelon form. But I am glad I got it somewhat right when saying it is independent of $x$. Thanks for your help
$endgroup$
– Nullspace
Jan 20 at 12:50
$begingroup$
@JackMoody I meant to say row echelon form. But I am glad I got it somewhat right when saying it is independent of $x$. Thanks for your help
$endgroup$
– Nullspace
Jan 20 at 12:50
1
1
$begingroup$
I would say that all you have done is OK to me.
$endgroup$
– idriskameni
Jan 20 at 12:52
$begingroup$
I would say that all you have done is OK to me.
$endgroup$
– idriskameni
Jan 20 at 12:52
1
1
$begingroup$
Where did $x$ go in your echelon form?
$endgroup$
– Bernard
Jan 20 at 13:20
$begingroup$
Where did $x$ go in your echelon form?
$endgroup$
– Bernard
Jan 20 at 13:20
1
1
$begingroup$
You're welcome. It's a pleasure to help!
$endgroup$
– Bernard
Jan 20 at 14:29
$begingroup$
You're welcome. It's a pleasure to help!
$endgroup$
– Bernard
Jan 20 at 14:29
|
show 8 more comments
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$begingroup$
This isn't reduced row echelon form. If you simplify it, you will find that you end up getting the identity matrix as your reduced echelon form. The takeaway should be that the rank, nullspace, and image are independent of x in this case.
$endgroup$
– Jack Moody
Jan 20 at 12:42
1
$begingroup$
@JackMoody I meant to say row echelon form. But I am glad I got it somewhat right when saying it is independent of $x$. Thanks for your help
$endgroup$
– Nullspace
Jan 20 at 12:50
1
$begingroup$
I would say that all you have done is OK to me.
$endgroup$
– idriskameni
Jan 20 at 12:52
1
$begingroup$
Where did $x$ go in your echelon form?
$endgroup$
– Bernard
Jan 20 at 13:20
1
$begingroup$
You're welcome. It's a pleasure to help!
$endgroup$
– Bernard
Jan 20 at 14:29