ans =
Does [V,D] = eig(A) always return normalized eigenvectors for any real matrix A?
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I’m looking for clarification on the eig function in MATLAB. When I use the syntax [V, D] = eig(A) for a real matrix A, are the eigenvectors returned in the columns of V guaranteed to be normalized to unit magnitude ( 
)?
)?Does this behavior change if the matrix is non-symmetric or if I use the generalized form eig(A, B) or in any other case for a real matrix A?
Thanks!
4 Comments
A = gallery("circul",3)
[V,D] = eig(A)
sqrt(V(:,1).^2 + V(:,2).^2 + V(:,3).^2)
So, NO, the magnitude of the second and third columns are 0.
I think the intent of Ali's phrase "normalized to unit magnitude" was asking more about this result from your example (regardless of his actual use of nomenclature):
A = gallery("circul",3)
[V,D] = eig(A)
sqrt(sum(conj(V).*V))
At least that is how I read his question.
A = gallery("circul",3);
[V,D] = eig(A);
arrayfun(@(i)norm(V(:,i)),1:width(V))
The individual eigenvectors are, per the documentation, however....
"[V,D] = eig(A) returns matrix V, whose columns are the right eigenvectors of A such that A*V = V*D. The eigenvectors in V are normalized so that the 2-norm of each is 1."
"If A is real symmetric, Hermitian, or skew-Hermitian, then the right eigenvectors V are orthonormal."
I think the above is true although I don't know that can prove it; probably somebody can illustrate a failing case.
David Goodmanson
1 minute ago
Hi dpb,
For the 'ortho' part of orthonomal he only 'failing' cases occur for two or more eigenvectors that have identical (degenerate) eigenvalues. These need not be mutually perpendicular but can be made so by constructing a mutually perpendicular set with appropriate linear combinations. For distinct eigenvalues, the eigenvectors are always mutually perpendicular for the real symmetric, hermitian or skew hermitian cases.
The 'normal' part is less fundamental and depends on how eig is called, as has been discussed. If there is any doubt on normalization, suppose you have m column eigenvectors each of length n that constitute an nxm matrix A (m<=n so not all of the eigenvectors are necessarily included). If you are confident that they are at least mutually orthogonal, you can use
n = sqrt(sum(A.*conj(A)));
Anew = A./n;
to get an orthonormal set.
Answers (3)
Paul
on 17 Mar 2026 at 20:35
1 vote
James Tursa
on 17 Mar 2026 at 20:39
1 vote
A discussion of eigenvector normalization can be found in the doc. In some cases, the answer is yes they are "normalized to unit magnitude" as you put it. But the answer can be no dependeing on the inputs.
1 Comment
Ali Baig
on 17 Mar 2026 at 23:04
John D'Errico
on 18 Mar 2026 at 13:06
Edited: John D'Errico
on 18 Mar 2026 at 13:08
Be a little careful that A is not in symbolic form, even if it is real and SPD.
A = randn(4,3); A = A'*A % A MUST clearly be SPD
[V,D] = eig(A)
It is true the columns of V are unit normalized when A is single or double precision.
diag(V'*V)
However, the symbolic version of A will not produce the same unit 2-normalized vectors.
[Vs,Ds] = eig(sym(A));
vpa(diag(Vs'*Vs))
6 Comments
James Tursa
on 18 Mar 2026 at 19:22
Edited: James Tursa
on 18 Mar 2026 at 19:38
Side Note on this calculation:
A = A'*A % A MUST clearly be SPD
The reason this is true numerically, as well as mathematically, is that MATLAB uses a BLAS symmetric matrix multiply routine in the background that enforces an exact symmetric result down to the least significant bit for this particular syntax. I.e., it only does about 1/2 of the calculations and then copies the results to get the exact numerical symmetry.
E.g., the following calculation would call a BLAS generic matrix multiply routine in the background, which is not guaranteed to generate an exact symmetric result:
AT = A';
A = AT * A; % A not guaranteed to be exactly symmetric
Paul
9 minutes ago
Any idea how Matlab handles
A.' * A
Does Matlab check if A is real and use the BLAS symmetric matrix multiply in that case and use the generic matrix multiply if A is complex?
James Tursa
about 2 hours ago
Edited: James Tursa
26 minutes ago
For A.' * A
Fact: BLAS library assumes interleaved storage for complex variables. To see routines available and interfaces you can go to this site (LAPACK and other library information is here also):
MATLAB always has to check if A is real to do the calculation. Once that is known, it will definitely call a BLAS real symmetric matrix multiply routine in the background (e.g., ssyrk or dsyrk) and produce an exact symmetric result if A is real. For complex A, I haven't checked which routine it will call. There are complex generic and symmetric and Hermitian routines available. For sure, it will not call the complex Hermitian routine (e.g., cherk or zherk) in the background of course. Whether it is smart enough to call the complex symmetric routine (e.g., csyrk or zsyrk) to save computation time and guarantee exact symmetric result with data copies or not, I don't know. I would hope so, but I simply haven't checked this stuff in detail since MATLAB went to interleaved complex storage.
E.g., consider the timings on this:
A = rand(7000) + rand(7000)*i;
AT = A.';
tic
ATA1 = A.'*A;
toc
tic
ATA2 = AT*A;
toc
isequal(ATA1,ATA2)
isequal(ATA1.',ATA1)
The timings suggest that different routines were called in the background. I would have to construct a mex routine that links with the BLAS library to confirm exactly which routines were called in the background, but the simple test above certainly supports the theory that the BLAS complex symmetric routine was called in the background for the first case.
SIDE NOTE: Back in the separate complex storage days pre R2018a, MATLAB would have to repeatedly call the BLAS real routines in the background with the real and imaginary pieces and combine the individual results to deal with complex matrix multiplies. But for R2018a onwards, since the background storage is interleaved it can call the BLAS complex matrix multiply routines directly. I'm confident MATLAB uses the Hermitian routines in the background when appropriate. I just haven't double checked the complex symmetric case (i.e., straight transpose instead of conjugate transpose) even though I would be surprised if they didn't take advantage of the complex symmetric routines and guarantee exact symmetric result via data copies.
If two different routines were called, is it surprising that they yielded the exact same result? I thought that one advantage of the ATA1 method is to ensure exact symmetry, but that seems to also obtain with ATA2 method.
Test case with a larger input.
rng(100);
A = rand(10000) + rand(10000)*i;
AT = A.';
tic
ATA1 = A.'*A;
toc
tic
ATA2 = AT*A;
toc
isequal(ATA1,ATA2)
isequal(ATA1.',ATA1)
James Tursa
13 minutes ago
Edited: James Tursa
9 minutes ago
I can't confirm this, but I believe this has changed slightly over the years in an attempt to make the two routines yield the same results even though one routine does about twice the actual calculations of the other. Earlier versions of MATLAB, I can confirm, do not necessarily yield exact symmetric results for generic matrix multiplies. I'm pretty sure there are some older posts of mine that demonstrate this. Whether this is still the case with recent versions of MATLAB is anyone's guess since I don't think the doc comments on it. Regardless, I wouldn't rely on it in any code I wrote, particularly if I wanted that code to port to older versions of MATLAB.
Quick test with non-square matrix:
A = rand(5000,4000);
AT = A.';
B = AT * A;
isequal(B.',B)
But on my R2021a Win64 PC:
>> A = rand(5000,4000);
>> AT = A.';
>> B = AT * A;
>> isequal(B.',B)
ans =
logical
0
Draw your own conclusions ...
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