Vector Spaces of Polynomials and Matrices, linear algebra I lesson

The polynomials $t^2$ and $-t^2 + t$ both have degree two, yet their sum is $t$ , which has degree one. Did adding two vectors take us out of the space? The answer depends on which set of polynomials we chose. This lesson states the rules that make a set with addition and scaling a vector space, checks them for polynomials of degree at most two and for 2-by-2 matrices, and gives these new vectors coordinates.

Key terms

Vector space

Definition A real vector space is a set $V$ with two operations. Addition assigns to any $\mathbf{u}$ and $\mathbf{v}$ in $V$ an object $\mathbf{u} + \mathbf{v}$ , and scalar multiplication assigns to any real number $a$ and any $\mathbf{u}$ in $V$ an object $a\mathbf{u}$ . The vector space axioms below must hold for all $\mathbf{u}, \mathbf{v}, \mathbf{w}$ in $V$ and all real numbers $a$ and $b$ .

The sum $\mathbf{u} + \mathbf{v}$ lies in $V$ (closure under addition).
The multiple $a\mathbf{u}$ lies in $V$ (closure under scalar multiplication).
Addition is commutative, $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$ .
Addition is associative, $(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$ .
There is a zero vector $\mathbf{0}$ in $V$ with $\mathbf{u} + \mathbf{0} = \mathbf{u}$ for every $\mathbf{u}$ .
Each $\mathbf{u}$ has an additive inverse $-\mathbf{u}$ in $V$ with $\mathbf{u} + (-\mathbf{u}) = \mathbf{0}$ .
Scalar multiplication is associative, $a(b\mathbf{u}) = (ab)\mathbf{u}$ .
Multiplying by one changes nothing, $1\mathbf{u} = \mathbf{u}$ .
A scalar distributes over a sum of vectors, $a(\mathbf{u} + \mathbf{v}) = a\mathbf{u} + a\mathbf{v}$ .
A sum of scalars distributes over a vector, $(a + b)\mathbf{u} = a\mathbf{u} + b\mathbf{u}$ .

The members of $V$ are called vectors, whatever kind of object they are. Linear combinations, span, linear independence, bases and coordinates are defined in $V$ word for word as in $\mathbb{R}^n$ . If $V$ has a basis of $n$ vectors, then every basis of $V$ has $n$ vectors, and $n$ is the dimension $\dim V$ . A subset of $V$ that contains $\mathbf{0}$ and is closed under addition and scalar multiplication is a subspace, and with the same operations it is a vector space in its own right.

In plain words A vector is anything you can add and scale by the familiar rules. Arrows are one example, but the word asks only that the two operations behave as listed.

Example On this page a tuple written inline, such as $(3, 2)$ , stands for a column vector. In $\mathbb{R}^2$ with the usual operations, every axiom comes down to real arithmetic in each component. For instance, $a(\mathbf{u} + \mathbf{v}) = a\mathbf{u} + a\mathbf{v}$ holds because $a(u_i + v_i) = au_i + av_i$ for $i = 1, 2$ . By contrast, the set of all vectors $(x, 1)$ with the same operations is not a vector space. The sum $(1, 1) + (2, 1) = (3, 2)$ falls outside the set, and so does $0 \cdot (1, 1) = (0, 0)$ . Both closure axioms fail, and the zero vector is missing.

Polynomials of degree at most two

Definition The set $P_2$ consists of all polynomials

p(t) = a_0 + a_1 t + a_2 t^2

with real coefficients $a_0, a_1, a_2$ , any of which may be zero. It contains the constants, the linear polynomials and the zero polynomial, whose coefficients are all zero. Two members are equal exactly when their corresponding coefficients are equal. For $q(t) = b_0 + b_1 t + b_2 t^2$ and a real number $c$ , addition and scalar multiplication act on the coefficients,

\begin{aligned} (p + q)(t) &= (a_0 + b_0) + (a_1 + b_1)t \\[0.4em] &\quad + (a_2 + b_2)t^2, \\[0.4em] (cp)(t) &= ca_0 + ca_1 t + ca_2 t^2, \end{aligned}

which is the same as adding or scaling the values at every input $t$ . No power above $t^2$ appears, so both operations stay in $P_2$ . Every other axiom follows coefficient by coefficient from the same law for real numbers. The zero vector is the zero polynomial, and the additive inverse of $p$ has every coefficient negated. With the same operations, the set of polynomials of degree exactly two is not a vector space.

In plain words The words “at most” do real work. They let the leading coefficients cancel, so sums and multiples never leave the set.

Example Take $p(t) = 2 + 3t - t^2$ and $q(t) = 1 - t + 2t^2$ . Adding and scaling coefficients gives

\begin{aligned} p + q &= 3 + 2t + t^2, \\[0.4em] -2p &= -4 - 6t + 2t^2, \end{aligned}

and both lie in $P_2$ . Now take $t^2$ and $-t^2 + t$ , which have degree exactly two. Their sum is $t$ , of degree one, and the multiple $0 \cdot t^2$ is the zero polynomial. Both results lie in $P_2$ but outside the set of polynomials of degree exactly two, so that set fails both closure axioms.

Coordinates of a polynomial

Definition The ordered list $\mathcal{S} = (1, t, t^2)$ is a basis of $P_2$ . It spans, since every member has the form $a_0 \cdot 1 + a_1 t + a_2 t^2$ , and it is linearly independent, since $a_0 + a_1 t + a_2 t^2$ is the zero polynomial only when $a_0 = a_1 = a_2 = 0$ . Hence $\dim P_2 = 3$ . If $\mathcal{B} = (p_1, p_2, p_3)$ is any ordered basis of $P_2$ , each $p$ in $P_2$ can be written in exactly one way as $p = c_1 p_1 + c_2 p_2 + c_3 p_3$ . The coordinates of $p$ in $\mathcal{B}$ form the column

[p]_{\mathcal{B}} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}.

In the basis $\mathcal{S}$ the coordinates are the coefficients, so $[a_0 + a_1 t + a_2 t^2]_{\mathcal{S}} = (a_0, a_1, a_2)$ .

In plain words The polynomial is the vector, and the column is its description in one chosen basis. Changing the basis can change the column, but it never changes the polynomial. The graph of $p$ is drawn in a plane, but $P_2$ has three independent coefficient directions, so the dimension of a picture and the dimension of the space are different questions.

Example The polynomial $p(t) = 2 + 3t - t^2$ has $[p]_{\mathcal{S}} = (2, 3, -1)$ . Now use $\mathcal{B} = (1, t - 1, (t - 1)^2)$ . Expanding a general combination gives

\begin{aligned} &\alpha + \beta(t - 1) + \gamma(t - 1)^2 \\[0.4em] &\quad = (\alpha - \beta + \gamma) \\[0.4em] &\qquad + (\beta - 2\gamma)t + \gamma t^2. \end{aligned}

Matching these coefficients with any target $a_0 + a_1 t + a_2 t^2$ fixes $\gamma$ , then $\beta$ , then $\alpha$ , with exactly one solution, so $\mathcal{B}$ is a basis. For our $p$ we get $\gamma = -1$ , then $\beta - 2\gamma = 3$ gives $\beta = 1$ , and $\alpha - \beta + \gamma = 2$ gives $\alpha = 4$ . So $[p]_{\mathcal{B}} = (4, 1, -1)$ . Reconstructing confirms it,

\begin{aligned} &4 + (t - 1) - (t - 1)^2 \\[0.4em] &\quad = 4 + t - 1 - t^2 + 2t - 1 \\[0.4em] &\quad = 2 + 3t - t^2. \end{aligned}

Space of 2-by-2 matrices

Definition The set $M_{2 \times 2}$ consists of all real $2 \times 2$ matrices. Addition adds corresponding entries, and scalar multiplication multiplies every entry by the scalar. With these operations $M_{2 \times 2}$ is a vector space, because each axiom holds entry by entry for real numbers. The zero vector is the zero matrix, and the additive inverse of a matrix negates all four entries. Matrix multiplication plays no part in this structure. In the same way the real $m \times n$ matrices form a vector space $M_{m \times n}$ . Matrices of different sizes cannot be added entrywise, so one space holds matrices of one fixed size.

In plain words Here a matrix is not acting on anything. It is a single vector, an array of four numbers that you add and scale one entry at a time.

Example Let

A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} -2 & 5 \\ 1 & 0 \end{bmatrix}.

Then

A + B = \begin{bmatrix} 0 & 4 \\ 4 & 4 \end{bmatrix}, \quad 2A = \begin{bmatrix} 4 & -2 \\ 6 & 8 \end{bmatrix}.

The axiom $A + B = B + A$ holds entry by entry. It says nothing about the products $AB$ and $BA$ , which belong to a different operation and here are not equal.

Matrix units

Definition The matrix units of $M_{2 \times 2}$ are

\begin{aligned} E_{11} &= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, & E_{12} &= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \\[0.4em] E_{21} &= \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, & E_{22} &= \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}. \end{aligned}

Each $E_{ij}$ has a $1$ in row $i$ and column $j$ and zeros elsewhere. In the order $(E_{11}, E_{12}, E_{21}, E_{22})$ they form an ordered basis of $M_{2 \times 2}$ , because every matrix satisfies

\begin{aligned} \begin{bmatrix} a & b \\ c & d \end{bmatrix} &= aE_{11} + bE_{12} \\[0.4em] &\quad + cE_{21} + dE_{22}, \end{aligned}

and this combination is the zero matrix only when $a = b = c = d = 0$ . Hence $\dim M_{2 \times 2} = 4$ , and the coordinate column of the matrix above is $(a, b, c, d)$ . Likewise the $mn$ matrix units $E_{ij}$ with $1 \le i \le m$ and $1 \le j \le n$ form a basis of $M_{m \times n}$ , so $\dim M_{m \times n} = mn$ .

In plain words Each matrix unit switches on one entry. The dimension counts the entries that can vary independently, four here, not the number of rows.

Example The matrix $A$ above is

A = 2E_{11} - E_{12} + 3E_{21} + 4E_{22},

so its coordinate column in this ordered basis is $(2, -1, 3, 4)$ . That column has four entries. It is not a column of $A$ , and it has nothing to do with $A$ multiplying a vector of $\mathbb{R}^2$ . It describes the single vector $A$ in the four-dimensional space $M_{2 \times 2}$ .

Check yourself

Which of these subsets of $P_2$ are subspaces? (a) The polynomials with constant coefficient $0$ . (b) The polynomials with constant coefficient $1$ . For each subspace, give a basis and the dimension.

Show the answer

Set (a) is a subspace. It contains the zero polynomial, and adding or scaling polynomials of the form $bt + ct^2$ keeps the constant coefficient equal to $0$ . Every member is a combination of $t$ and $t^2$ , and $bt + ct^2$ is the zero polynomial only when $b = c = 0$ . So $(t, t^2)$ is a basis and the dimension is $2$ .

Set (b) is not a subspace. The zero polynomial has constant coefficient $0$ , so it is missing. Closure under addition fails too, since $(1 + t) + (1 + t^2) = 2 + t + t^2$ has constant coefficient $2$ .

Show that $\mathcal{B} = (1, 1 + t, t^2)$ is a basis of $P_2$ , and find the coordinates of $p(t) = 2 + 3t - t^2$ in it.

Show the answer

A general combination is

\begin{aligned} &\alpha + \beta(1 + t) + \gamma t^2 \\[0.4em] &\quad = (\alpha + \beta) + \beta t + \gamma t^2. \end{aligned}

It equals $a + bt + ct^2$ exactly when $\beta = b$ , $\gamma = c$ and $\alpha = a - b$ . Every target has exactly one solution, so $\mathcal{B}$ spans $P_2$ , and the zero polynomial comes only from $\alpha = \beta = \gamma = 0$ , so $\mathcal{B}$ is independent. For $p$ we get $\beta = 3$ , $\gamma = -1$ and $\alpha = 2 - 3 = -1$ , so $[p]_{\mathcal{B}} = (-1, 3, -1)$ . Check that $-1 + 3(1 + t) - t^2 = 2 + 3t - t^2$ . The polynomial is unchanged, and only its coordinate column differs from $(2, 3, -1)$ .

A $2 \times 2$ matrix is symmetric when its two off-diagonal entries are equal. Show that the symmetric matrices form a subspace of $M_{2 \times 2}$ and find a basis and the dimension. Is $(E_{11}, E_{12}, E_{22})$ a basis of this subspace?

Show the answer

The zero matrix is symmetric. Adding two symmetric matrices gives equal off-diagonal entries $b + b'$ , and multiplying by $k$ gives equal entries $kb$ , so the set is a subspace. Every symmetric matrix can be written as

\begin{aligned} \begin{bmatrix} a & b \\ b & c \end{bmatrix} &= aE_{11} + b(E_{12} + E_{21}) \\[0.4em] &\quad + cE_{22}. \end{aligned}

The three matrices $E_{11}$ , $E_{12} + E_{21}$ and $E_{22}$ are symmetric and span the subspace. They are independent, because the $(1,1)$ , $(1,2)$ and $(2,2)$ entries of the combination are $a$ , $b$ and $c$ . So the dimension is $3$ .

The list $(E_{11}, E_{12}, E_{22})$ is not a basis. Its middle matrix has off-diagonal entries $1$ and $0$ , so it is not symmetric and does not even lie in the subspace.

Vector Spaces of Polynomials and Matrices

Key terms

Vector space

Polynomials of degree at most two

Coordinates of a polynomial

Space of 2-by-2 matrices

Matrix units

Check yourself

How this lesson connects

Builds on

Used later in