Proving the existence of a complex one dimensional subspace of non invertible matrices
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Let $n$ be an integer $geq2$ and let $A,Bin M_{ntimes n}(mathbb{C})$ such that A and B are invertible and the set ${A,B}$ is linearly independent. Prove that there exists a one-dimensional subspace $W$ of $span({A,B})$ such that no element of $W$ is invertible.
Essentially, all I have to do is show that I can create a non invertible matrix from a linear combination from $A$ and $B$. This is because any multiple of a non invertible matrix will result in a non invertible matrix. Currently, what I'm trying to show is that there exists $k_1,k_2in mathbb{C}$ such that there exists $c_1,c_2,c_3in mathbb{C}$ not all $0$ such that $$c_1(k_1a_1+k_2b_1)+c_2(k_1a_2+k_2b_2)+c_3(k_1a_3+k_2b_3)=0$$ where $a_1,a_2,a_3$ and $b_1,b_2,b_3$ are the row vectors of $A$ and $B$ respectively. I am not sure where to go from here.
linear-algebra
edited Nov 14 at 8:58
Yadati Kiran
358111
asked Nov 14 at 8:11
Ryan Greyling
564
add a comment |
up vote
0
down vote
favorite
Let $n$ be an integer $geq2$ and let $A,Bin M_{ntimes n}(mathbb{C})$ such that A and B are invertible and the set ${A,B}$ is linearly independent. Prove that there exists a one-dimensional subspace $W$ of $span({A,B})$ such that no element of $W$ is invertible.
Essentially, all I have to do is show that I can create a non invertible matrix from a linear combination from $A$ and $B$. This is because any multiple of a non invertible matrix will result in a non invertible matrix. Currently, what I'm trying to show is that there exists $k_1,k_2in mathbb{C}$ such that there exists $c_1,c_2,c_3in mathbb{C}$ not all $0$ such that $$c_1(k_1a_1+k_2b_1)+c_2(k_1a_2+k_2b_2)+c_3(k_1a_3+k_2b_3)=0$$ where $a_1,a_2,a_3$ and $b_1,b_2,b_3$ are the row vectors of $A$ and $B$ respectively. I am not sure where to go from here.
linear-algebra
edited Nov 14 at 8:58
Yadati Kiran
358111
asked Nov 14 at 8:11
Ryan Greyling
564
Are $a_1,a_2,a_3$ row vectors or column vectors?
– Yadati Kiran
Nov 14 at 8:33
@YadatiKiran They are row vectors.
– Ryan Greyling
Nov 14 at 14:43
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $n$ be an integer $geq2$ and let $A,Bin M_{ntimes n}(mathbb{C})$ such that A and B are invertible and the set ${A,B}$ is linearly independent. Prove that there exists a one-dimensional subspace $W$ of $span({A,B})$ such that no element of $W$ is invertible.
Essentially, all I have to do is show that I can create a non invertible matrix from a linear combination from $A$ and $B$. This is because any multiple of a non invertible matrix will result in a non invertible matrix. Currently, what I'm trying to show is that there exists $k_1,k_2in mathbb{C}$ such that there exists $c_1,c_2,c_3in mathbb{C}$ not all $0$ such that $$c_1(k_1a_1+k_2b_1)+c_2(k_1a_2+k_2b_2)+c_3(k_1a_3+k_2b_3)=0$$ where $a_1,a_2,a_3$ and $b_1,b_2,b_3$ are the row vectors of $A$ and $B$ respectively. I am not sure where to go from here.
linear-algebra
edited Nov 14 at 8:58
Yadati Kiran
358111
asked Nov 14 at 8:11
Ryan Greyling
564
Let $n$ be an integer $geq2$ and let $A,Bin M_{ntimes n}(mathbb{C})$ such that A and B are invertible and the set ${A,B}$ is linearly independent. Prove that there exists a one-dimensional subspace $W$ of $span({A,B})$ such that no element of $W$ is invertible.
Essentially, all I have to do is show that I can create a non invertible matrix from a linear combination from $A$ and $B$. This is because any multiple of a non invertible matrix will result in a non invertible matrix. Currently, what I'm trying to show is that there exists $k_1,k_2in mathbb{C}$ such that there exists $c_1,c_2,c_3in mathbb{C}$ not all $0$ such that $$c_1(k_1a_1+k_2b_1)+c_2(k_1a_2+k_2b_2)+c_3(k_1a_3+k_2b_3)=0$$ where $a_1,a_2,a_3$ and $b_1,b_2,b_3$ are the row vectors of $A$ and $B$ respectively. I am not sure where to go from here.
linear-algebra
linear-algebra
edited Nov 14 at 8:58
Yadati Kiran
358111
asked Nov 14 at 8:11
Ryan Greyling
564
edited Nov 14 at 8:58
Yadati Kiran
358111
asked Nov 14 at 8:11
Ryan Greyling
564
edited Nov 14 at 8:58
Yadati Kiran
358111
edited Nov 14 at 8:58
Yadati Kiran
358111
edited Nov 14 at 8:58
Yadati Kiran
358111
358111
asked Nov 14 at 8:11
Ryan Greyling
564
asked Nov 14 at 8:11
Ryan Greyling
564
asked Nov 14 at 8:11
Ryan Greyling
564
564
Are $a_1,a_2,a_3$ row vectors or column vectors?
– Yadati Kiran
Nov 14 at 8:33
@YadatiKiran They are row vectors.
– Ryan Greyling
Nov 14 at 14:43
add a comment |
Are $a_1,a_2,a_3$ row vectors or column vectors?
– Yadati Kiran
Nov 14 at 8:33
@YadatiKiran They are row vectors.
– Ryan Greyling
Nov 14 at 14:43
Are $a_1,a_2,a_3$ row vectors or column vectors?
– Yadati Kiran
Nov 14 at 8:33
Are $a_1,a_2,a_3$ row vectors or column vectors?
– Yadati Kiran
Nov 14 at 8:33
@YadatiKiran They are row vectors.
– Ryan Greyling
Nov 14 at 14:43
@YadatiKiran They are row vectors.
– Ryan Greyling
Nov 14 at 14:43
add a comment |
1 Answer
1
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oldest
votes
up vote
2
down vote
accepted
Just pick an eigenvalue $lambda$ of $A^{-1}B$ and consider the linear span of $lambda A-B$.
answered Nov 14 at 8:40
user1551
70k566125
Yes, from my testing that appears to be correct. I'm just having trouble understanding why.
– Ryan Greyling
Nov 14 at 14:53
@RyanGreyling $lambda A-B=A(lambda I-A^{-1}B)$.
– user1551
Nov 14 at 16:36
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Just pick an eigenvalue $lambda$ of $A^{-1}B$ and consider the linear span of $lambda A-B$.
answered Nov 14 at 8:40
user1551
70k566125
Yes, from my testing that appears to be correct. I'm just having trouble understanding why.
– Ryan Greyling
Nov 14 at 14:53
@RyanGreyling $lambda A-B=A(lambda I-A^{-1}B)$.
– user1551
Nov 14 at 16:36
add a comment |
up vote
2
down vote
accepted
Just pick an eigenvalue $lambda$ of $A^{-1}B$ and consider the linear span of $lambda A-B$.
answered Nov 14 at 8:40
user1551
70k566125
Yes, from my testing that appears to be correct. I'm just having trouble understanding why.
– Ryan Greyling
Nov 14 at 14:53
@RyanGreyling $lambda A-B=A(lambda I-A^{-1}B)$.
– user1551
Nov 14 at 16:36
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Just pick an eigenvalue $lambda$ of $A^{-1}B$ and consider the linear span of $lambda A-B$.
answered Nov 14 at 8:40
user1551
70k566125
Just pick an eigenvalue $lambda$ of $A^{-1}B$ and consider the linear span of $lambda A-B$.
answered Nov 14 at 8:40
user1551
70k566125
answered Nov 14 at 8:40
user1551
70k566125
answered Nov 14 at 8:40
user1551
70k566125
answered Nov 14 at 8:40
user1551
70k566125
70k566125
Yes, from my testing that appears to be correct. I'm just having trouble understanding why.
– Ryan Greyling
Nov 14 at 14:53
@RyanGreyling $lambda A-B=A(lambda I-A^{-1}B)$.
– user1551
Nov 14 at 16:36
add a comment |
Yes, from my testing that appears to be correct. I'm just having trouble understanding why.
– Ryan Greyling
Nov 14 at 14:53
@RyanGreyling $lambda A-B=A(lambda I-A^{-1}B)$.
– user1551
Nov 14 at 16:36
Yes, from my testing that appears to be correct. I'm just having trouble understanding why.
– Ryan Greyling
Nov 14 at 14:53
Yes, from my testing that appears to be correct. I'm just having trouble understanding why.
– Ryan Greyling
Nov 14 at 14:53
@RyanGreyling $lambda A-B=A(lambda I-A^{-1}B)$.
– user1551
Nov 14 at 16:36
@RyanGreyling $lambda A-B=A(lambda I-A^{-1}B)$.
– user1551
Nov 14 at 16:36
add a comment |
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Are $a_1,a_2,a_3$ row vectors or column vectors?
– Yadati Kiran
Nov 14 at 8:33
@YadatiKiran They are row vectors.
– Ryan Greyling
Nov 14 at 14:43