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Section5.6Abstract Vector Spaces

Up to this point the only vector spaces we have considered are vector spaces where the vectors are tuples of numbers, i.e. \(\R^n\) where \(n=1,2,3, \ldots\text{,}\) but many other kinds of vector spaces exist and are very useful. In fact, vector spaces are quite ubiquitous throughout mathematics.

SubsectionMatrix Vector Spaces

Example5.6.1Matrix Vector Spaces

Consider the set of \(m\times n\) matrices with real entries which we will denote, \(M_{mn}(\R)\text{.}\) This set is a vector space. To see why let \(A, B\) be elements of \(M_{mn}(\R)\text{,}\) and let \(c\) be any real number then

  1. \(M_{mn}(\R)\) is not empty, specifically it contains an \(m\times n\) matrix made of all zeros which serves as our zero vector.

  2. This set is closed under addition, \(A+B \in M_{mn}(\R)\text{.}\)

  3. This set is closed under scalar multiplication, \(cA \in M_{mn}(\R)\text{.}\)

  4. Matrix addition is commutative, \(A+B = B+A\) for all matrices in \(M_{mn}(\R)\text{.}\)

More concretely, consider \(M_{22}(\R)\text{,}\) the set of \(2\times 2\) matrices with real entries. What subset of \(2\times 2\text{,}\) is a basis for this vector space? Well, whatever it is it must allow us to write any matrix as a linear combination of its elements. The simplest choice is called the standard basis, and in this case the simplest choice is the set: \begin{equation*} \mathcal{B} = \left\{ \begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \end{bmatrix}, \begin{bmatrix} 0 \amp 1 \\ 0 \amp 0 \end{bmatrix}, \begin{bmatrix} 0 \amp 0 \\ 1 \amp 0 \end{bmatrix}, \begin{bmatrix} 0 \amp 0 \\ 0 \amp 1 \end{bmatrix} \right\} \end{equation*} This allow us to write for example, \begin{equation*} \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix} = a\begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \end{bmatrix} + b\begin{bmatrix} 0 \amp 1 \\ 0 \amp 0 \end{bmatrix} + c\begin{bmatrix} 0 \amp 0 \\ 1 \amp 0 \end{bmatrix} + d\begin{bmatrix} 0 \amp 0 \\ 0 \amp 1 \end{bmatrix} \end{equation*} Is \(\mathcal{B}\) really a basis? Clearly it spans the vector space \(M_{22}(\R)\text{,}\) but is there perhaps a smaller set which still spans \(M_{22}(\R)\text{?}\) Also, how do we know that the four matrices in \(\mathcal{B}\) are linearly independent?

SubsectionSolution Space of Linear, Homogeneous DEs

Example5.6.2Solution Space of Homogeneous, Linear DEs

Consider the differential equation \begin{equation} y'' + y = 0. \label{eqn-sho}\tag{5.6.1} \end{equation} You can check that both \begin{align*} y_1 \amp = \cos x, \quad \text{ and}\\ y_2 \amp = \sin x \end{align*} are solutions. But also, any linear combination of these two solutions is again a solution. To see this let \(y = ay_1 + by_2\) where \(a\) and \(b\) are scalars, then: \begin{align*} y \amp = a\cos x + b\sin x\\ y' \amp = -a\sin x + b\cos x\\ y'' \amp = -a\cos x - b\sin x \end{align*} So \(y'' + y = (-a\cos x - b\sin x) + (a\cos x + b\sin x) = 0\text{,}\) and thus we see that the set of solutions is nonempty and closed under linear combinations and therefore a vector space.

Notice that if equation (5.6.1) were not homogeneous, that is if the right hand side of the equation were not zero, then the set of solutions would not form a vector space.

Example5.6.3Solution Set of Nonhomogeneous DEs

Consider the differential equation, \begin{equation} y'' + y = e^x. \label{eqn-forced-sho}\tag{5.6.2} \end{equation} You can check that both \begin{align*} y_1 \amp = \cos x + \frac{1}{2}e^x, \quad \text{ and}\\ y_2 \amp = \sin x + \frac{1}{2}e^x \end{align*} are solutions. However, linear combinations of these two solutions are not solutions. To see this let \(y = ay_1 + by_2\) where \(a\) and \(b\) are scalars, then: \begin{align*} y \amp = a\left(\cos x + \frac{1}{2}e^x \right) + b\left(\sin x + \frac{1}{2}e^x \right)\\ y' \amp = a\left(-\sin x + \frac{1}{2}e^x \right) + b\left(\cos x + \frac{1}{2}e^x \right)\\ y'' \amp = a\left(-\cos x + \frac{1}{2}e^x \right) + b\left(-\sin x + \frac{1}{2}e^x \right) \end{align*} \begin{align*} y'' + y \amp = a\frac{1}{2}e^x + b\frac{1}{2}e^x + a\frac{1}{2}e^x + b\frac{1}{2}e^x\\ \amp = (a+b) e^x \end{align*} Thus \(y_1\) and \(y_2\) will only be solutions when \(a+b=1\text{.}\) In other words, the solutions to equation (5.6.2) do not form a vector subspace.