Mixing
If $A$ and $B$ are square matrices of order $n$, which of these statements are true?
$begingroup$
If $AB = B$, then $B$ is the identity matrix.
If matrix $A$ is reversible, then $ABA^{−1} = B$
Are these two equations true? Do square matrices have commutativity property? Can you explain me this? I'm beginner in math.
linear-algebra matrices
edited Jan 30 at 17:45
stressed out
6,5131939
asked Jan 30 at 17:43
ElyEly
101
$endgroup$
add a comment |
$begingroup$
If $AB = B$, then $B$ is the identity matrix.
If matrix $A$ is reversible, then $ABA^{−1} = B$
Are these two equations true? Do square matrices have commutativity property? Can you explain me this? I'm beginner in math.
linear-algebra matrices
edited Jan 30 at 17:45
stressed out
6,5131939
asked Jan 30 at 17:43
ElyEly
101
$endgroup$
add a comment |
$begingroup$
If $AB = B$, then $B$ is the identity matrix.
If matrix $A$ is reversible, then $ABA^{−1} = B$
Are these two equations true? Do square matrices have commutativity property? Can you explain me this? I'm beginner in math.
linear-algebra matrices
edited Jan 30 at 17:45
stressed out
6,5131939
asked Jan 30 at 17:43
ElyEly
101
$endgroup$
If $AB = B$, then $B$ is the identity matrix.
If matrix $A$ is reversible, then $ABA^{−1} = B$
Are these two equations true? Do square matrices have commutativity property? Can you explain me this? I'm beginner in math.
linear-algebra matrices
linear-algebra matrices
edited Jan 30 at 17:45
stressed out
6,5131939
asked Jan 30 at 17:43
ElyEly
101
edited Jan 30 at 17:45
stressed out
6,5131939
asked Jan 30 at 17:43
ElyEly
101
edited Jan 30 at 17:45
stressed out
6,5131939
edited Jan 30 at 17:45
stressed out
6,5131939
edited Jan 30 at 17:45
stressed out
6,5131939
6,5131939
asked Jan 30 at 17:43
ElyEly
101
asked Jan 30 at 17:43
ElyEly
101
asked Jan 30 at 17:43
ElyEly
101
101
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Neither is true. Consider $A=I$. Then $AB=B$ for any $B$.
For $2$, just consider a nondiagonal matrix which is diagonalizable.
One of the first things you will discover is that matrix multiplication isn't commutative.
For $2$, consider $begin{pmatrix}1&1\0&1end{pmatrix}$ and $begin{pmatrix}0&1\1&0end{pmatrix}$.
Try finding a different example.
answered Jan 30 at 17:59
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
Given the level of the question, I think OP might not know what "diagonalizable" means.
$endgroup$
– Wojowu
Jan 30 at 18:12
$begingroup$
True. It's the first example that came to mind. I guess I'll try to produce a simpler one.
$endgroup$
– Chris Custer
Jan 30 at 18:17
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Neither is true. Consider $A=I$. Then $AB=B$ for any $B$.
For $2$, just consider a nondiagonal matrix which is diagonalizable.
One of the first things you will discover is that matrix multiplication isn't commutative.
For $2$, consider $begin{pmatrix}1&1\0&1end{pmatrix}$ and $begin{pmatrix}0&1\1&0end{pmatrix}$.
Try finding a different example.
answered Jan 30 at 17:59
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
Given the level of the question, I think OP might not know what "diagonalizable" means.
$endgroup$
– Wojowu
Jan 30 at 18:12
$begingroup$
True. It's the first example that came to mind. I guess I'll try to produce a simpler one.
$endgroup$
– Chris Custer
Jan 30 at 18:17
add a comment |
$begingroup$
Neither is true. Consider $A=I$. Then $AB=B$ for any $B$.
For $2$, just consider a nondiagonal matrix which is diagonalizable.
One of the first things you will discover is that matrix multiplication isn't commutative.
For $2$, consider $begin{pmatrix}1&1\0&1end{pmatrix}$ and $begin{pmatrix}0&1\1&0end{pmatrix}$.
Try finding a different example.
answered Jan 30 at 17:59
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
Given the level of the question, I think OP might not know what "diagonalizable" means.
$endgroup$
– Wojowu
Jan 30 at 18:12
$begingroup$
True. It's the first example that came to mind. I guess I'll try to produce a simpler one.
$endgroup$
– Chris Custer
Jan 30 at 18:17
add a comment |
$begingroup$
Neither is true. Consider $A=I$. Then $AB=B$ for any $B$.
For $2$, just consider a nondiagonal matrix which is diagonalizable.
One of the first things you will discover is that matrix multiplication isn't commutative.
For $2$, consider $begin{pmatrix}1&1\0&1end{pmatrix}$ and $begin{pmatrix}0&1\1&0end{pmatrix}$.
Try finding a different example.
answered Jan 30 at 17:59
Chris CusterChris Custer
14.3k3827
$endgroup$
Neither is true. Consider $A=I$. Then $AB=B$ for any $B$.
For $2$, just consider a nondiagonal matrix which is diagonalizable.
One of the first things you will discover is that matrix multiplication isn't commutative.
For $2$, consider $begin{pmatrix}1&1\0&1end{pmatrix}$ and $begin{pmatrix}0&1\1&0end{pmatrix}$.
Try finding a different example.
answered Jan 30 at 17:59
Chris CusterChris Custer
14.3k3827
edited Jan 30 at 19:05
answered Jan 30 at 17:59
Chris CusterChris Custer
14.3k3827
answered Jan 30 at 17:59
Chris CusterChris Custer
14.3k3827
answered Jan 30 at 17:59
Chris CusterChris Custer
14.3k3827
14.3k3827
$begingroup$
Given the level of the question, I think OP might not know what "diagonalizable" means.
$endgroup$
– Wojowu
Jan 30 at 18:12
$begingroup$
True. It's the first example that came to mind. I guess I'll try to produce a simpler one.
$endgroup$
– Chris Custer
Jan 30 at 18:17
add a comment |
$begingroup$
Given the level of the question, I think OP might not know what "diagonalizable" means.
$endgroup$
– Wojowu
Jan 30 at 18:12
$begingroup$
True. It's the first example that came to mind. I guess I'll try to produce a simpler one.
$endgroup$
– Chris Custer
Jan 30 at 18:17
$begingroup$
Given the level of the question, I think OP might not know what "diagonalizable" means.
$endgroup$
– Wojowu
Jan 30 at 18:12
$begingroup$
Given the level of the question, I think OP might not know what "diagonalizable" means.
$endgroup$
– Wojowu
Jan 30 at 18:12
$begingroup$
True. It's the first example that came to mind. I guess I'll try to produce a simpler one.
$endgroup$
– Chris Custer
Jan 30 at 18:17
$begingroup$
True. It's the first example that came to mind. I guess I'll try to produce a simpler one.
$endgroup$
– Chris Custer
Jan 30 at 18:17
add a comment |
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