Can eigenvectors be scaled and still be eigenvectors?Given the matrix: A = [ 2 0 1 4 ] .The eigenvalues are: λ 1 = 2 , λ 2 = 4.To find the eigenvectors: v 1 = nullity ⁡ ( A − λ 1 I ) , v 2 = nullity ⁡ ( A − λ 2 I ) .Which yielded the following eigenvectors for v 1 : v 1 = [ − 2 3 1 ] .However, I have seen this alternative eigenvector for v 1 : v 1 = [ − 2 3 ] .Question: Are both eigenvectors for v 1 correct? In other words, can you scale an eigenvector by any real number and it still be an eigenvector of that matrix?

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Can eigenvectors be scaled and still be eigenvectors?Given the matrix:  A  =      [                            2                          0                                      1                          4                      ]    .The eigenvalues are:            λ        1    =  2  ,            λ        2    =  4.To find the eigenvectors:      v    1    =  nullity  ⁡  (  A  −            λ        1    I  )  ,      v    2    =  nullity  ⁡  (  A  −            λ        2    I  )  .Which yielded the following eigenvectors for       v    1  :      v    1    =      [                            −                      2            3                                                1                      ]    .However, I have seen this alternative eigenvector for       v    1  :      v    1    =      [                            −          2                                      3                      ]    .Question: Are both eigenvectors for       v    1   correct? In other words, can you scale an eigenvector by any real number and it still be an eigenvector of that matrix?

Avah Leonard · Accepted Answer

A formal proof of this statement could be the following: Let A be a square matrix with eigenvector   v and eigenvalue   λ (i.e.   A  v  =  λ  v). We would like to show that cv=x is also an eigenvector, where   c  ≠  0 is a constant. This follows from  A  x  =  A  (  c  v  )  =  c  (  A  v  )  =  c  (  λ  v  )  =  λ  (  c  v  )  =  λ  x  .Note that the case c=0 has to be excluded since x=0 is by definition never an eigenvector.

Jenny Stafford · Accepted Answer

To add to first&#039;s answer, the eigenvectors of a linear map corresponding to a certain eigenvalue form a subspace of their own. This is because linear combinations of eigenvectors with the same eigenvalue produce more eigenvectors:Let             v      i       be eigenvectors with eigenvalue   λ. Then   T  (  ∑            a      i              v      i        )  =  ∑            a      i        T    (          v      i        )    =  ∑            a      i        λ          v      i        =  λ  ∑            a      i              v      i      . This shows that the linear combination is an eigenvector with eigenvalue   λ

Given the matrix: [[2,0],[1,4]] The eigenvalues are: lambda_1=2, lambda_2=4. To find the eigenvectors: v_1=nullity(A−lambda_1I), v_2=nullity(A−lambda_2I).

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