Dimension of subspace 10x10
Question:
Let $W$ be the subspace of a $M_{10,10}$ consisting of all matrices whose diagonal entries are zero. Find the dimension of $W$.
Not really sure how to approach this. Any suggestions?
linear-algebra
asked Nov 28 at 19:43
Forextrader
346
add a comment |
Question:
Let $W$ be the subspace of a $M_{10,10}$ consisting of all matrices whose diagonal entries are zero. Find the dimension of $W$.
Not really sure how to approach this. Any suggestions?
linear-algebra
asked Nov 28 at 19:43
Forextrader
346
1
$$10x10-10=100-10=90$${}{}{}{}
– hamam_Abdallah
Nov 28 at 19:49
Have you learnt the Rank-Nullity theorem yet?
– PersonX
Nov 28 at 19:58
add a comment |
Question:
Let $W$ be the subspace of a $M_{10,10}$ consisting of all matrices whose diagonal entries are zero. Find the dimension of $W$.
Not really sure how to approach this. Any suggestions?
linear-algebra
asked Nov 28 at 19:43
Forextrader
346
Question:
Let $W$ be the subspace of a $M_{10,10}$ consisting of all matrices whose diagonal entries are zero. Find the dimension of $W$.
Not really sure how to approach this. Any suggestions?
linear-algebra
linear-algebra
asked Nov 28 at 19:43
Forextrader
346
asked Nov 28 at 19:43
Forextrader
346
asked Nov 28 at 19:43
Forextrader
346
asked Nov 28 at 19:43
Forextrader
346
asked Nov 28 at 19:43
Forextrader
346
346
1
$$10x10-10=100-10=90$${}{}{}{}
– hamam_Abdallah
Nov 28 at 19:49
Have you learnt the Rank-Nullity theorem yet?
– PersonX
Nov 28 at 19:58
add a comment |
1
$$10x10-10=100-10=90$${}{}{}{}
– hamam_Abdallah
Nov 28 at 19:49
Have you learnt the Rank-Nullity theorem yet?
– PersonX
Nov 28 at 19:58
1
1
$$10x10-10=100-10=90$${}{}{}{}
– hamam_Abdallah
Nov 28 at 19:49
$$10x10-10=100-10=90$${}{}{}{}
– hamam_Abdallah
Nov 28 at 19:49
Have you learnt the Rank-Nullity theorem yet?
– PersonX
Nov 28 at 19:58
Have you learnt the Rank-Nullity theorem yet?
– PersonX
Nov 28 at 19:58
add a comment |
2 Answers
2
active
oldest
votes
Intuitively, the dimension of $M_{10,10}$ is $100$ because you need to specify all $100$ entries to determine a matrix uniquely. All the matrices in $W$ have zeros on the diagonal, so we need only specify the other $90$ non-diagonal entries. Therefore, the dimension is $90$.
I admit this argument is not completely rigorous but hopefully will give you the intuition needed to write a rigorous proof.
answered Nov 28 at 19:47
pwerth
1,500411
add a comment |
Hint:
Consider the linear map $;begin{aligned}[t]f:mathcal M_{10times 10}&longrightarrow K^{10}, \ (a_{ij})_{1le i,jle 10}&longmapsto(a_{ii})_{1le ile 10} end{aligned}$
Observe that $f$ is onto and $W=ker f$.
answered Nov 28 at 20:04
Bernard
117k638111
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Intuitively, the dimension of $M_{10,10}$ is $100$ because you need to specify all $100$ entries to determine a matrix uniquely. All the matrices in $W$ have zeros on the diagonal, so we need only specify the other $90$ non-diagonal entries. Therefore, the dimension is $90$.
I admit this argument is not completely rigorous but hopefully will give you the intuition needed to write a rigorous proof.
answered Nov 28 at 19:47
pwerth
1,500411
add a comment |
Intuitively, the dimension of $M_{10,10}$ is $100$ because you need to specify all $100$ entries to determine a matrix uniquely. All the matrices in $W$ have zeros on the diagonal, so we need only specify the other $90$ non-diagonal entries. Therefore, the dimension is $90$.
I admit this argument is not completely rigorous but hopefully will give you the intuition needed to write a rigorous proof.
answered Nov 28 at 19:47
pwerth
1,500411
add a comment |
Intuitively, the dimension of $M_{10,10}$ is $100$ because you need to specify all $100$ entries to determine a matrix uniquely. All the matrices in $W$ have zeros on the diagonal, so we need only specify the other $90$ non-diagonal entries. Therefore, the dimension is $90$.
I admit this argument is not completely rigorous but hopefully will give you the intuition needed to write a rigorous proof.
answered Nov 28 at 19:47
pwerth
1,500411
Intuitively, the dimension of $M_{10,10}$ is $100$ because you need to specify all $100$ entries to determine a matrix uniquely. All the matrices in $W$ have zeros on the diagonal, so we need only specify the other $90$ non-diagonal entries. Therefore, the dimension is $90$.
I admit this argument is not completely rigorous but hopefully will give you the intuition needed to write a rigorous proof.
answered Nov 28 at 19:47
pwerth
1,500411
answered Nov 28 at 19:47
pwerth
1,500411
answered Nov 28 at 19:47
pwerth
1,500411
answered Nov 28 at 19:47
pwerth
1,500411
1,500411
add a comment |
add a comment |
Hint:
Consider the linear map $;begin{aligned}[t]f:mathcal M_{10times 10}&longrightarrow K^{10}, \ (a_{ij})_{1le i,jle 10}&longmapsto(a_{ii})_{1le ile 10} end{aligned}$
Observe that $f$ is onto and $W=ker f$.
answered Nov 28 at 20:04
Bernard
117k638111
add a comment |
Hint:
Consider the linear map $;begin{aligned}[t]f:mathcal M_{10times 10}&longrightarrow K^{10}, \ (a_{ij})_{1le i,jle 10}&longmapsto(a_{ii})_{1le ile 10} end{aligned}$
Observe that $f$ is onto and $W=ker f$.
answered Nov 28 at 20:04
Bernard
117k638111
add a comment |
Hint:
Consider the linear map $;begin{aligned}[t]f:mathcal M_{10times 10}&longrightarrow K^{10}, \ (a_{ij})_{1le i,jle 10}&longmapsto(a_{ii})_{1le ile 10} end{aligned}$
Observe that $f$ is onto and $W=ker f$.
answered Nov 28 at 20:04
Bernard
117k638111
Hint:
Consider the linear map $;begin{aligned}[t]f:mathcal M_{10times 10}&longrightarrow K^{10}, \ (a_{ij})_{1le i,jle 10}&longmapsto(a_{ii})_{1le ile 10} end{aligned}$
Observe that $f$ is onto and $W=ker f$.
answered Nov 28 at 20:04
Bernard
117k638111
answered Nov 28 at 20:04
Bernard
117k638111
answered Nov 28 at 20:04
Bernard
117k638111
answered Nov 28 at 20:04
Bernard
117k638111
117k638111
add a comment |
add a comment |
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1
$$10x10-10=100-10=90$${}{}{}{}
– hamam_Abdallah
Nov 28 at 19:49
Have you learnt the Rank-Nullity theorem yet?
– PersonX
Nov 28 at 19:58