Bases, operators and eigenvectors

[fog37]

Hello,

In the case of 2D vector spaces, every vector member of the vector space can be expressed as a linear combination of two independent vectors which together form a basis. There are infinitely many possible and valid bases, each containing two independent vectors (not necessarily orthogonal or orthonormal), that we can use. Vectors existing in a 3D vector space can use one out of the many and infinitely possible sets of 3 independent base vectors and so on...

When talking about linear operators and their eigenvectors, we learn that the set of eigenvectors (each eigenvector has its own eigenvalue) of an operator forms a base since the eigenvectors are independent vectors.
That means that the base formed by the eigenvectors of a certain linear operator is just one of those many, infinite possible bases, correct? In 2D, linear operators must therefore have only two eigenvectors (forming a 2D base) while in 3D linear operators can only have 3 eigenvectors and so... is that true?

If we considered a generic basis for a 2D vector space, would the vectors in that basis necessarily be the two eigenvectors of some linear operator? Or is that not necessarily true?

Thanks,
Fog37

 

[BvU]

In 2D, linear operators must therefore have only two eigenvectors (forming a 2D base) while in 3D linear operators can only have 3 eigenvectors and so... is that true?

Counter-example: $\vec v\rightarrow\ 3\vec v$
(you want different eigenvalues)

even then: if $\vec a$ is an eigenvector, then any multiple of $\vec a$ is also an eigenvector
(you want independent eigenvectors)

would the vectors in that basis necessarily be the two eigenvectors

You can always construct such an operator: assign (different) eigenvalues to these eigenvectors and there you are !

 

[lavinia]

Be aware that a linear operator may not have any eigenvectors.

 

[fresh_42]

Check out
$$
\begin{bmatrix}1&0\\0&1 \end{bmatrix}\; , \;\begin{bmatrix}1&0\\0&-1 \end{bmatrix}\; , \;\begin{bmatrix}1&1\\0&1 \end{bmatrix}\, , \,\begin{bmatrix}0&-1\\1&0 \end{bmatrix}\; , \;\begin{bmatrix}1&1\\0&0 \end{bmatrix}
$$

 

[Stephen Tashi]

When talking about linear operators and their eigenvectors, we learn that the set of eigenvectors (each eigenvector has its own eigenvalue) of an operator forms a base since the eigenvectors are independent vectors.

Perhaps what you have in mind is the situation where a linear operator that maps an n-dimensional space into itself has n distinct eigenvalues and n-distinct eigenvectors. This is not the situation that applies to all linear operators, but there are particular examples of this situation that are important in physics. You hear more about these particular operators than other operators, so you can get the impression that all operators are this way.

This about an operator that projects 3D space onto a 1D subspace - for example T((x,y,z)) = (x,0,0).

 

[fog37]

Thanks everyone. All very helpful. Thank you for reminding me that any scaled version of the same eigenvector is also an eigenvector. So a base can be formed by eigenvectors having different eigenvalues which makes them independent (orthogonal).

In summary:
In 2D, for example, we can form infinite new bases from one particular basis of two independent vectors by scaling each vector in the basis by arbitrary amounts. BvU mentioned that, given an arbitrary basis of two orthogonal vectors, we can find/construct a linear operator that has the two orthogonal vectors as its eigenvectors.

Of course, an acceptable basis is also one that has two vectors that are independent but not orthogonal. In this case, I don't think we can arrive to a linear operator having those independent vectors as eigenvectors. Are bases having non-orthogonal independent vectors very useful? If so, do you have an example?

 

[BvU]

independent (orthogonal).

Note that independent $\ne$ orthogonal at all !

 

[fog37]

yes, two vectors, graphically, are independent as long as they are not along the same line of action. their mutual angle can be any angle other than zero or 180.

Still, independent vectors form a basis...