Mixing
The work of elimination is cut in half when $A$ is symmetric
$begingroup$
I am going through the book "Gilbert Strang - Introduction to Linear Algebra, Third Edition (2003)" and as I was working through the chapter of transpose matrices and solving systems of linear equations, I got stuck at this one particular sentence.
What does the author mean by 'smaller matrices', and how do they stay the same as elimination proceeds.
I tried doing the process by hand, chose a 3x3 symmetric matrix, and I didn't see the symmetry until the very end: $A = LDU = LDL^T$
Also, as you may notice, in the end he even says the work of elimination is cut in half. I really don't see this.
linear-algebra
asked Feb 3 at 9:00
zafirzaryazafirzarya
424
$endgroup$
add a comment |
$begingroup$
I am going through the book "Gilbert Strang - Introduction to Linear Algebra, Third Edition (2003)" and as I was working through the chapter of transpose matrices and solving systems of linear equations, I got stuck at this one particular sentence.
What does the author mean by 'smaller matrices', and how do they stay the same as elimination proceeds.
I tried doing the process by hand, chose a 3x3 symmetric matrix, and I didn't see the symmetry until the very end: $A = LDU = LDL^T$
Also, as you may notice, in the end he even says the work of elimination is cut in half. I really don't see this.
linear-algebra
asked Feb 3 at 9:00
zafirzaryazafirzarya
424
$endgroup$
2
$begingroup$
After one step of Gaussian elimination on an $n times n$ matrix $M_{n}$, you get a matrix of the form $$pmatrix{ a_{1} & a_{2} & ldots & a_{n} \ 0 & & & \ vdots & & M_{n-1}& \ 0 & & & \ }$$ where $M_{n-1}$ is an $(n-1) times (n-1)$ (smaller) matrix. I think these are the `smaller matrices' which are all symmetric.
$endgroup$
– preferred_anon
Feb 3 at 9:38
1
$begingroup$
Yes, I see it now. Thanks a ton
$endgroup$
– zafirzarya
Feb 3 at 9:51
add a comment |
$begingroup$
I am going through the book "Gilbert Strang - Introduction to Linear Algebra, Third Edition (2003)" and as I was working through the chapter of transpose matrices and solving systems of linear equations, I got stuck at this one particular sentence.
What does the author mean by 'smaller matrices', and how do they stay the same as elimination proceeds.
I tried doing the process by hand, chose a 3x3 symmetric matrix, and I didn't see the symmetry until the very end: $A = LDU = LDL^T$
Also, as you may notice, in the end he even says the work of elimination is cut in half. I really don't see this.
linear-algebra
asked Feb 3 at 9:00
zafirzaryazafirzarya
424
$endgroup$
I am going through the book "Gilbert Strang - Introduction to Linear Algebra, Third Edition (2003)" and as I was working through the chapter of transpose matrices and solving systems of linear equations, I got stuck at this one particular sentence.
What does the author mean by 'smaller matrices', and how do they stay the same as elimination proceeds.
I tried doing the process by hand, chose a 3x3 symmetric matrix, and I didn't see the symmetry until the very end: $A = LDU = LDL^T$
Also, as you may notice, in the end he even says the work of elimination is cut in half. I really don't see this.
linear-algebra
linear-algebra
asked Feb 3 at 9:00
zafirzaryazafirzarya
424
asked Feb 3 at 9:00
zafirzaryazafirzarya
424
asked Feb 3 at 9:00
zafirzaryazafirzarya
424
asked Feb 3 at 9:00
zafirzaryazafirzarya
424
asked Feb 3 at 9:00
zafirzaryazafirzarya
424
424
2
$begingroup$
After one step of Gaussian elimination on an $n times n$ matrix $M_{n}$, you get a matrix of the form $$pmatrix{ a_{1} & a_{2} & ldots & a_{n} \ 0 & & & \ vdots & & M_{n-1}& \ 0 & & & \ }$$ where $M_{n-1}$ is an $(n-1) times (n-1)$ (smaller) matrix. I think these are the `smaller matrices' which are all symmetric.
$endgroup$
– preferred_anon
Feb 3 at 9:38
1
$begingroup$
Yes, I see it now. Thanks a ton
$endgroup$
– zafirzarya
Feb 3 at 9:51
add a comment |
2
$begingroup$
After one step of Gaussian elimination on an $n times n$ matrix $M_{n}$, you get a matrix of the form $$pmatrix{ a_{1} & a_{2} & ldots & a_{n} \ 0 & & & \ vdots & & M_{n-1}& \ 0 & & & \ }$$ where $M_{n-1}$ is an $(n-1) times (n-1)$ (smaller) matrix. I think these are the `smaller matrices' which are all symmetric.
$endgroup$
– preferred_anon
Feb 3 at 9:38
1
$begingroup$
Yes, I see it now. Thanks a ton
$endgroup$
– zafirzarya
Feb 3 at 9:51
2
2
$begingroup$
After one step of Gaussian elimination on an $n times n$ matrix $M_{n}$, you get a matrix of the form $$pmatrix{ a_{1} & a_{2} & ldots & a_{n} \ 0 & & & \ vdots & & M_{n-1}& \ 0 & & & \ }$$ where $M_{n-1}$ is an $(n-1) times (n-1)$ (smaller) matrix. I think these are the `smaller matrices' which are all symmetric.
$endgroup$
– preferred_anon
Feb 3 at 9:38
$begingroup$
After one step of Gaussian elimination on an $n times n$ matrix $M_{n}$, you get a matrix of the form $$pmatrix{ a_{1} & a_{2} & ldots & a_{n} \ 0 & & & \ vdots & & M_{n-1}& \ 0 & & & \ }$$ where $M_{n-1}$ is an $(n-1) times (n-1)$ (smaller) matrix. I think these are the `smaller matrices' which are all symmetric.
$endgroup$
– preferred_anon
Feb 3 at 9:38
1
1
$begingroup$
Yes, I see it now. Thanks a ton
$endgroup$
– zafirzarya
Feb 3 at 9:51
$begingroup$
Yes, I see it now. Thanks a ton
$endgroup$
– zafirzarya
Feb 3 at 9:51
add a comment |
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2
$begingroup$
After one step of Gaussian elimination on an $n times n$ matrix $M_{n}$, you get a matrix of the form $$pmatrix{ a_{1} & a_{2} & ldots & a_{n} \ 0 & & & \ vdots & & M_{n-1}& \ 0 & & & \ }$$ where $M_{n-1}$ is an $(n-1) times (n-1)$ (smaller) matrix. I think these are the `smaller matrices' which are all symmetric.
$endgroup$
– preferred_anon
Feb 3 at 9:38
1
$begingroup$
Yes, I see it now. Thanks a ton
$endgroup$
– zafirzarya
Feb 3 at 9:51