Let us now use the symmetric and antisymmetric tensor products to define two subspaces of the tensor square 
Elements of 
Elements of
All of what we have said above can be generalized in a natural way to products of more than two vectors. More precisely, for any natural number 
of 
for all 
for all 
for all 
Even better, we can use the scalar product so defined to construct an orthonormal basis of 
where 
In particular, since the cardinality of 
Example 1: If 
We now define the
Reading Assignment: Familiarize yourself with permutations. What is important for our purposes is that you understand how to multiply permutations, and that you understand what the sign of a permutation is. Feel free to ask questions as needed.
Definition 1: For any 
and denote by 
and denote by 
Note that, in the case 
and
from Lecture 21.
Since the symmetric and antisymmetric tensor products are defined in terms of the tensor product, they inherit multilinearity. For example, in the case 
for all
for all
for all
The symmetric tensor product is constructed in such a way that
for any permutation 
for any permutation 
Indeed, suppose that 
but on the other hand we also have
because
if 
The vector space 
Theorem 1: For any 
and
Since we won’t use this theorem much, we will skip the proof. However, the proof is not too difficult, and is an exercise in permutations: simply plug in the definitions of the symmetric and antisymmetric tensor products in terms of the original tensor products, expand the scalar product, and simplify.
Perhaps counterintuitively, the antisymmetric tensor product is more important than the symmetric tensor product in linear algebra. The next theorem explains why.
Theorem 2: For any 
Proof: Suppose first that 
If 
We then have that
by multinearity of the wedge product. Now observe that the 
which contains the vector 
Conversely, suppose 
We must prove that
We prove this by induction on 
For the inductive step, we proceed as follows. Since
Let 
by
We then have that
Now, since 
It then follows that
since otherwise the linear transformation 
— Q.E.D.
Corollary 1: We have 
Since
you can think of this as a massive generalization of the Cauchy-Schwarz inequality, which is the case
One one hand, the scalar product is a proper multiplication rule in the sense that it satisfies the 
In this case, vectors 
are real numbers, and the vector product 
is the product of real numbers.
Technically, we have not seen this example yet, but here it is. Let 
and 
be vectors in 
Next week, we will see that this example of vector multiplication gives the complex number system.
In this example, the vector space 
which are identically zero after finitely many terms. This means that 
be vectors in 
which is just the recipe for multiplying polynomials and collecting together terms of the same degree.
In this example, the vector space 
we associate a new vector denoted 
which is called the tensor product of 
and 
However, the vector 
does not reside in 
What is happening here is that we view the symbol 
as a rule for multiplying two vectors, but we do not specify what this rule is — instead, we view 
which contains all unevaluated products of vectors from 
is a natural number and 
are vectors. These unevaluated expressions are called 
are equal if and only if it is possible to rewrite 
as 
using bilinearity.
we have
produces the zero vector 
This was proved in 
is essentially the same (try it!).
in the vector space 
we have 
Using bilinearity, we have 
now follows from the case 
Then, there is an associated scalar product on the vector space 
is an orthogonal set of vectors in 
are different tensors, then their scalar product is zero. We have
, and the second implies 
In either case, the product 
is equal to zero.
is an orthonormal basis in 
is an orthonormal basis in 
spans 
is an orthogonal set in 
then 
However, we can decompose a simple tensor into two pieces, as
as a symmetric version of the tensor product: a bilinear multiplication of vectors that, by construction, is commutative:
the symmetric tensor product produces the same tensor as the tensor product itself:
Let 
remaining at the end of this computation as a 
is said to be selfadjoint (or symmetric) if 
The goal of this lecture is to prove the following cornerstone result in linear algebra.
one vector at a time. A nice feature of the construction is that the eigenbasis it outputs is an orthonormal basis of 
is said to be nonnegative if the associated quadratic form is nonnegative, i.e. if the function defined by 
for all 
has the property that membership in its kernel is certified by vanishing of 
if and only if 
then 
and let 
be any number and let 
be an arbitrary vector. We have
together with the fact that 
this further simplifies to 
this parabola is upward=opening. Moreover, since the lefthand side satisfies 
the lowest point of this parabola cannot lie below the line 
and this forces 
was chosen arbitrarily, so the above equation holds for any 
in particular 
We thus have
i.e. 
; the argument even gives a description of the corresponding eigenvalue 
The quadratic form 
is a continuous function, and hence by the 
on the sphere,
at which the minimum is achieved, i.e.
of 
and the minimizer 
we have that 
for any 
the above inequality can be rewritten as 
is nonnegative. Moreover, we have that 
meaning that
be a linear operator, and let 
if 
then 
i.e. as an element of the algebra 
Consider the orthogonal complement of the line spanned by 
i.e. the subspace of 
is invariant under 
of 
This follows easily from the fact that 
-dimensional subspace 
But this means that we can simply apply Theorem 2 again, with 
where 
and 
is a vector at which the minimum is achieved,
is a unit vector orthogonal to 
is a linearly independent set in 
since 
is a subset of 
as a kind of 
; these are the 
where 
is the 
in the algebra 
meaning that it is the 
and more generally if 
is any polynomial in a single variable then we have
even works in the case that 
is a negative integer, provided that 
equivalently, the simple operator 
is invertible if and only if 
its inverse being 
If 
and 
are simple operators, then they commute,
in two variables. ![[A]_E = \begin{bmatrix} \lambda & {} & {} \\ {} & \ddots & {} \\ {} & {} & \lambda \end{bmatrix},](https://s0.wp.com/latex.php?latex=%5BA%5D_E+%3D+%5Cbegin%7Bbmatrix%7D+%5Clambda+%26+%7B%7D+%26+%7B%7D+%5C%5C+%7B%7D+%26+%5Cddots+%26+%7B%7D+%5C%5C+%7B%7D+%26+%7B%7D+%26+%5Clambda+%5Cend%7Bbmatrix%7D%2C&bg=FFFFFF&fg=000&s=0&c=20201002 "[A]_E = \begin{bmatrix} \lambda & {} & {} \\ {} & \ddots & {} \\ {} & {} & \lambda \end{bmatrix},")
we look for special vectors in 
is said to be an eigenvector of an operator 
if 
The scalar 
is said to be semisimple if there exists a basis 
such that 
then ![[A]_E = \begin{bmatrix} \lambda_1 & {} & {} \\ {} & \ddots & {} \\ {} & {} & \lambda_n \end{bmatrix}.](https://s0.wp.com/latex.php?latex=%5BA%5D_E+%3D+%5Cbegin%7Bbmatrix%7D+%5Clambda_1+%26+%7B%7D+%26+%7B%7D+%5C%5C+%7B%7D+%26+%5Cddots+%26+%7B%7D+%5C%5C+%7B%7D+%26+%7B%7D+%26+%5Clambda_n+%5Cend%7Bbmatrix%7D.&bg=FFFFFF&fg=000&s=0&c=20201002 "[A]_E = \begin{bmatrix} \lambda_1 & {} & {} \\ {} & \ddots & {} \\ {} & {} & \lambda_n \end{bmatrix}.")
counterclockwise through the angle 
The matrix of this operator relative to the standard basis 
is 
then 
, so that 
is a simple operator: 
are eigenvectors, with eigenvalues 
If 
then 
and again 
However, taking any other value of 
for example 
rotation through a right angle, it is geometrically clear that 
is never a scalar multiple of 
so that 
defined by 
the set 
defined by
is called the geometric multiplicity of 
considered above has empty spectrum, 
the eigenspace 
because 
then 
then 
A key feature of these subspaces is that they are independent of one another.
are distinct eigenvalues of an operator 
Let 
be nonzero vectors such that 
for each 
Then 
is a linearly independent set.
The base case is 
consisting of a single eigenvector of 
not all equal to zero, such that 
Applying the operator 
is a linearly independent set, and hence all the coefficients in this vector equation are zero. In particular, we have
and 
Hence, the set 
Theorem 1 has the following crucial consequences.
be a basis of the eigenspace 
of 
then 
is equivalent to saying that the spectrum of 
distinct numbers,
of eigenvectors of 
is orthogonal if the image 
of any orthonormal basis 
of 
-element of the matrix ![[B]_E](https://s0.wp.com/latex.php?latex=%5BB%5D_E&bg=FFFFFF&fg=000&s=0&c=20201002 "[B]_E")
is equal to the 
-element of the matrix ![[A]_E,](https://s0.wp.com/latex.php?latex=%5BA%5D_E%2C&bg=FFFFFF&fg=000&s=0&c=20201002 "[A]_E,")
a relationship which is usually expressed as saying that ![[A]_E.](https://s0.wp.com/latex.php?latex=%5BA%5D_E.&bg=FFFFFF&fg=000&s=0&c=20201002 "[A]_E.")
Now, for any vectors 
are two operators such that
![[B-C]_E](https://s0.wp.com/latex.php?latex=%5BB-C%5D_E&bg=FFFFFF&fg=000&s=0&c=20201002 "[B-C]_E")
is the zero matrix. Since the map which takes an operator to its matrix relative to 
is the zero operator, or equivalently that 
the unique operator such that 
for all 
. We call 
is the inverse of 
i.e. 
Operators on Euclidean space which are related to their adjoint in some predictable way play a very important role in linear algebra. 
is said to be selfadjoint if 
.
satisfies 
and also 
If 
and also 
by choosing a basis 
be an orthonormal basis in 
Let 
be a linear transformation.
are the scalar products 
is called a “matrix element” of ![[A]_{E,F}](https://s0.wp.com/latex.php?latex=%5BA%5D_%7BE%2CF%7D&bg=FFFFFF&fg=000&s=0&c=20201002 "[A]_{E,F}")
of defined in 
However, one can note that it is not actually necessary to assume that ![[A]_{E,F} = \begin{bmatrix} {} & \vdots & {} \\ \dots & \langle \mathbf{f}_i,A\mathbf{e}_j \rangle & \dots \\ {} & \vdots & {} \end{bmatrix}_{1 \leq i \leq m, 1 \leq j \leq n}.](https://s0.wp.com/latex.php?latex=%5BA%5D_%7BE%2CF%7D+%3D+%5Cbegin%7Bbmatrix%7D+%7B%7D+%26+%5Cvdots+%26+%7B%7D+%5C%5C+%5Cdots+%26+%5Clangle+%5Cmathbf%7Bf%7D_i%2CA%5Cmathbf%7Be%7D_j+%5Crangle+%26+%5Cdots+%5C%5C+%7B%7D+%26+%5Cvdots+%26+%7B%7D+%5Cend%7Bbmatrix%7D_%7B1+%5Cleq+i+%5Cleq+m%2C+1+%5Cleq+j+%5Cleq+n%7D.&bg=FFFFFF&fg=000&s=0&c=20201002 "[A]_{E,F} = \begin{bmatrix} {} & \vdots & {} \\ \dots & \langle \mathbf{f}_i,A\mathbf{e}_j \rangle & \dots \\ {} & \vdots & {} \end{bmatrix}_{1 \leq i \leq m, 1 \leq j \leq n}.")
so that 
and 
are two (possibly) different orthonormal bases of the same Euclidean space. Given an operator 
matrices ![[A]_E \quad\text{ and }\quad [A]_F](https://s0.wp.com/latex.php?latex=%5BA%5D_E+%5Cquad%5Ctext%7B+and+%7D%5Cquad+%5BA%5D_F&bg=FFFFFF&fg=000&s=0&c=20201002 "[A]_E \quad\text{ and }\quad [A]_F")
respectively. In order to do this, let us consider the linear operator 
uniquely determined by the 
Operators which transform orthonormal bases into orthonormal bases have a special name.
in 
is again an orthonormal basis in 
we defined above by 
are orthogonal operators.
and 
for some 
where 
Then 
is an orthonormal basis of 
so that 
we have 
![[A]_E,[A]_F](https://s0.wp.com/latex.php?latex=%5BA%5D_E%2C%5BA%5D_F&bg=FFFFFF&fg=000&s=0&c=20201002 "[A]_E,[A]_F")
. We have 
![[A]_F = [UAU^{-1}]_E = [U]_E [A]_E [U^{-1}]_E = [U]_E [A]_E [U]_E^{-1}](https://s0.wp.com/latex.php?latex=%5BA%5D_F+%3D+%5BUAU%5E%7B-1%7D%5D_E+%3D+%5BU%5D_E+%5BA%5D_E+%5BU%5E%7B-1%7D%5D_E+%3D+%5BU%5D_E+%5BA%5D_E+%5BU%5D_E%5E%7B-1%7D&bg=FFFFFF&fg=000&s=0&c=20201002 "[A]_F = [UAU^{-1}]_E = [U]_E [A]_E [U^{-1}]_E = [U]_E [A]_E [U]_E^{-1}")
![[\cdot]_E \colon \mathrm{End}\mathrm{V} \to \mathbb{R}^{n\times n}](https://s0.wp.com/latex.php?latex=%5B%5Ccdot%5D_E+%5Ccolon+%5Cmathrm%7BEnd%7D%5Cmathrm%7BV%7D+%5Cto+%5Cmathbb%7BR%7D%5E%7Bn%5Ctimes+n%7D&bg=FFFFFF&fg=000&s=0&c=20201002 "[\cdot]_E \colon \mathrm{End}\mathrm{V} \to \mathbb{R}^{n\times n}")
![[A]_F](https://s0.wp.com/latex.php?latex=%5BA%5D_F&bg=FFFFFF&fg=000&s=0&c=20201002 "[A]_F")
representing the operator ![[A]_E](https://s0.wp.com/latex.php?latex=%5BA%5D_E&bg=FFFFFF&fg=000&s=0&c=20201002 "[A]_E")
representing ![[U]_E,](https://s0.wp.com/latex.php?latex=%5BU%5D_E%2C&bg=FFFFFF&fg=000&s=0&c=20201002 "[U]_E,")
where 
of linear transformations from a vector space 
of linear operators on 
This notation stems from the fact that a fancy name for a linear operator is 
we may define their product to be their composition, i.e. AB:=A \circ B. Spelled out, this means that 
is the linear operator defined by
is a special type of vector space whose vectors (which are operators) can be scaled, added, and multiplied. Such vector spaces warrant their own name.
and now we find that there are in fact many more algebras, namely all vector spaces 
such that 
where 
is the identity operator defined by 
for all 
being an isomorphism of 
then there is precisely one operator 
You should check for yourself that 
is invertible, and that its inverse is 
of invertible operators is closed under multiplication: if 
then 
is invertible, and that 
.
-dimensional vector space 
as a linear combination of the vectors in 
relative to the basis 
![[A\mathbf{v}]_F = [A]_{E,F} [\mathbf{v}]_E,](https://s0.wp.com/latex.php?latex=%5BA%5Cmathbf%7Bv%7D%5D_F+%3D+%5BA%5D_%7BE%2CF%7D+%5B%5Cmathbf%7Bv%7D%5D_E%2C&bg=FFFFFF&fg=000&s=0&c=20201002 "[A\mathbf{v}]_F = [A]_{E,F} [\mathbf{v}]_E,")
![[\mathbf{v}]_E](https://s0.wp.com/latex.php?latex=%5B%5Cmathbf%7Bv%7D%5D_E&bg=FFFFFF&fg=000&s=0&c=20201002 "[\mathbf{v}]_E")
denotes the 
matrix whose entries are the coordinates of the vector ![[\mathbf{w}]_F](https://s0.wp.com/latex.php?latex=%5B%5Cmathbf%7Bw%7D%5D_F&bg=FFFFFF&fg=000&s=0&c=20201002 "[\mathbf{w}]_F")
denotes the 
matrix whose entries are the coordinates of the vector 
relative to the basis 
matrix ![[A]_{E,F} = \begin{bmatrix} [A\mathbf{e}_1]_F & \dots & [A\mathbf{e}_n]_F \end{bmatrix}](https://s0.wp.com/latex.php?latex=%5BA%5D_%7BE%2CF%7D+%3D+%5Cbegin%7Bbmatrix%7D+%5BA%5Cmathbf%7Be%7D_1%5D_F+%26+%5Cdots+%26+%5BA%5Cmathbf%7Be%7D_n%5D_F+%5Cend%7Bbmatrix%7D&bg=FFFFFF&fg=000&s=0&c=20201002 "[A]_{E,F} = \begin{bmatrix} [A\mathbf{e}_1]_F & \dots & [A\mathbf{e}_n]_F \end{bmatrix}")
th column is the ![[A\mathbf{e}_j]_F.](https://s0.wp.com/latex.php?latex=%5BA%5Cmathbf%7Be%7D_j%5D_F.&bg=FFFFFF&fg=000&s=0&c=20201002 "[A\mathbf{e}_j]_F.")
What this means at the conceptual level is the following: choosing a basis 
defined by![\mathbf{v} \mapsto [\mathbf{v}]_E,](https://s0.wp.com/latex.php?latex=%5Cmathbf%7Bv%7D+%5Cmapsto+%5B%5Cmathbf%7Bv%7D%5D_E%2C&bg=FFFFFF&fg=000&s=0&c=20201002 "\mathbf{v} \mapsto [\mathbf{v}]_E,")
defined by![\mathbf{w} \mapsto [\mathbf{w}]_F,](https://s0.wp.com/latex.php?latex=%5Cmathbf%7Bw%7D+%5Cmapsto+%5B%5Cmathbf%7Bw%7D%5D_F%2C&bg=FFFFFF&fg=000&s=0&c=20201002 "\mathbf{w} \mapsto [\mathbf{w}]_F,")
defined by ![A \mapsto [A]_{E,F}.](https://s0.wp.com/latex.php?latex=A+%5Cmapsto+%5BA%5D_%7BE%2CF%7D.&bg=FFFFFF&fg=000&s=0&c=20201002 "A \mapsto [A]_{E,F}.")
and 
We are then dealing with linear operators ![[A]_E = \begin{bmatrix} [A\mathbf{e}_1]_E & \dots & [A\mathbf{e}_n]_E \end{bmatrix}.](https://s0.wp.com/latex.php?latex=%5BA%5D_E+%3D+%5Cbegin%7Bbmatrix%7D+%5BA%5Cmathbf%7Be%7D_1%5D_E+%26+%5Cdots+%26+%5BA%5Cmathbf%7Be%7D_n%5D_E+%5Cend%7Bbmatrix%7D.&bg=FFFFFF&fg=000&s=0&c=20201002 "[A]_E = \begin{bmatrix} [A\mathbf{e}_1]_E & \dots & [A\mathbf{e}_n]_E \end{bmatrix}.")
![[A\mathbf{v}]_E = [A]_E [\mathbf{v}]_E.](https://s0.wp.com/latex.php?latex=%5BA%5Cmathbf%7Bv%7D%5D_E+%3D+%5BA%5D_E+%5B%5Cmathbf%7Bv%7D%5D_E.&bg=FFFFFF&fg=000&s=0&c=20201002 "[A\mathbf{v}]_E = [A]_E [\mathbf{v}]_E.")
Then, we also have their product 
and a natural issue is to determine the relationship between the matrices ![[A]_E,[B]_E,](https://s0.wp.com/latex.php?latex=%5BA%5D_E%2C%5BB%5D_E%2C&bg=FFFFFF&fg=000&s=0&c=20201002 "[A]_E,[B]_E,")
and ![[AB]_E.](https://s0.wp.com/latex.php?latex=%5BAB%5D_E.&bg=FFFFFF&fg=000&s=0&c=20201002 "[AB]_E.")
Let us now work this out.
relative to the basis 
![[AB]_E = [A]_E [B]_E.](https://s0.wp.com/latex.php?latex=%5BAB%5D_E+%3D+%5BA%5D_E+%5BB%5D_E.&bg=FFFFFF&fg=000&s=0&c=20201002 "[AB]_E = [A]_E [B]_E.")
given by ![A \mapsto [A]_E](https://s0.wp.com/latex.php?latex=A+%5Cmapsto+%5BA%5D_E&bg=FFFFFF&fg=000&s=0&c=20201002 "A \mapsto [A]_E")
