Mixing
sum of two linearly independent eigenvector
0
$begingroup$
'' The sum of two linearly independent eigenvectors is never an eigenvector. ''
- This is true?
- if i assume the opposite, then it means the eigenvector can be written as a combination linear of the 2 other eigenvector (coefficient 1 of each), which won't make them linearly independent, which means they wouldn't be eigenvectors in the first place, is this correct?
linear-algebra
edited Jan 29 at 6:47
Bungo
13.7k22148
asked Jan 29 at 3:44
ximxim
516
$endgroup$
add a comment |
0
$begingroup$
'' The sum of two linearly independent eigenvectors is never an eigenvector. ''
- This is true?
- if i assume the opposite, then it means the eigenvector can be written as a combination linear of the 2 other eigenvector (coefficient 1 of each), which won't make them linearly independent, which means they wouldn't be eigenvectors in the first place, is this correct?
linear-algebra
edited Jan 29 at 6:47
Bungo
13.7k22148
asked Jan 29 at 3:44
ximxim
516
$endgroup$
$begingroup$
Rolled back to the original because the edit mangled the question so it no longer made sense.
$endgroup$
– Bungo
Jan 29 at 6:48
add a comment |
0
0
0
$begingroup$
'' The sum of two linearly independent eigenvectors is never an eigenvector. ''
- This is true?
- if i assume the opposite, then it means the eigenvector can be written as a combination linear of the 2 other eigenvector (coefficient 1 of each), which won't make them linearly independent, which means they wouldn't be eigenvectors in the first place, is this correct?
linear-algebra
edited Jan 29 at 6:47
Bungo
13.7k22148
asked Jan 29 at 3:44
ximxim
516
$endgroup$
'' The sum of two linearly independent eigenvectors is never an eigenvector. ''
- This is true?
- if i assume the opposite, then it means the eigenvector can be written as a combination linear of the 2 other eigenvector (coefficient 1 of each), which won't make them linearly independent, which means they wouldn't be eigenvectors in the first place, is this correct?
linear-algebra
linear-algebra
edited Jan 29 at 6:47
Bungo
13.7k22148
asked Jan 29 at 3:44
ximxim
516
edited Jan 29 at 6:47
Bungo
13.7k22148
asked Jan 29 at 3:44
ximxim
516
edited Jan 29 at 6:47
Bungo
13.7k22148
edited Jan 29 at 6:47
Bungo
13.7k22148
edited Jan 29 at 6:47
Bungo
13.7k22148
13.7k22148
asked Jan 29 at 3:44
ximxim
516
asked Jan 29 at 3:44
ximxim
516
asked Jan 29 at 3:44
ximxim
516
516
$begingroup$
Rolled back to the original because the edit mangled the question so it no longer made sense.
$endgroup$
– Bungo
Jan 29 at 6:48
add a comment |
$begingroup$
Rolled back to the original because the edit mangled the question so it no longer made sense.
$endgroup$
– Bungo
Jan 29 at 6:48
$begingroup$
Rolled back to the original because the edit mangled the question so it no longer made sense.
$endgroup$
– Bungo
Jan 29 at 6:48
$begingroup$
Rolled back to the original because the edit mangled the question so it no longer made sense.
$endgroup$
– Bungo
Jan 29 at 6:48
add a comment |
1 Answer
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Consider the identity matrix, every single non-zero vector of compatible size is an eigenvector.
Hence the statement is false.
answered Jan 29 at 3:48
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
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$begingroup$
Consider the identity matrix, every single non-zero vector of compatible size is an eigenvector.
Hence the statement is false.
answered Jan 29 at 3:48
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
1
$begingroup$
Consider the identity matrix, every single non-zero vector of compatible size is an eigenvector.
Hence the statement is false.
answered Jan 29 at 3:48
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
1
1
1
$begingroup$
Consider the identity matrix, every single non-zero vector of compatible size is an eigenvector.
Hence the statement is false.
answered Jan 29 at 3:48
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
Consider the identity matrix, every single non-zero vector of compatible size is an eigenvector.
Hence the statement is false.
answered Jan 29 at 3:48
Siong Thye GohSiong Thye Goh
103k1468119
edited Jan 29 at 6:04
answered Jan 29 at 3:48
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 29 at 3:48
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 29 at 3:48
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
add a comment |
add a comment |
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$begingroup$
Rolled back to the original because the edit mangled the question so it no longer made sense.
$endgroup$
– Bungo
Jan 29 at 6:48