3.1: Vector Spaces
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In multivariable, or vector calculus, a vector is defined to be a mathematical construct that has both direction and magnitude. In linear algebra, vectors are defined more abstractly. Vectors are mathematical constructs that can be added and multiplied by scalars under the usual rules of arithmetic. Vector addition is commutative and associative, and scalar multiplication is distributive and associative. Let \(\text{u}\), \(\text{v}\), and \(\text{w}\) be vectors, and let \(a\), \(b\), and \(c\) be scalars. Then the rules of arithmetic say that
\[\text{u} + \text{v} = \text{v} + \text{u},\quad \text{u} + (\text{v} + \text{w}) = (\text{u} + \text{v}) + \text{w};\nonumber \]
and
\[a(\text{u}+\text{v})=a\text{u}+a\text{v},\quad a(b\text{u})=(ab)\text{u}.\nonumber \]
A vector space consists of a set of vectors and a set of scalars that is closed under vector addition and scalar multiplication. That is, when you multiply any two vectors in a vector space by scalars and add them, the resulting vector is still in the vector space.
We can give some examples of vector spaces. Let the scalars be the set of real numbers and let the vectors be column matrices of a specified type. One example of a vector space is the set of all three-by-one column matrices. If we let
\[\text{u}=\left(\begin{array}{c}u_1\\u_2\\u_3\end{array}\right),\quad\text{v}=\left(\begin{array}{c}v_1\\v_2\\v_3\end{array}\right),\nonumber \]
then
\[w=a\text{u}+b\text{v}=\left(\begin{array}{c}au_1+bv_1 \\ au_2+bv_2 \\ au_3+bv_3\end{array}\right)\nonumber \]
is evidently a three-by-one matrix, so that the set of all three-by-one matrices (together with the set of real numbers) forms a vector space. This vector space is usually called \(\mathbb{R}^3\).
A vector subspace is a vector space that is a subset of another vector space. For example, a vector subspace of \(\mathbb{R}^3\) could be the set of all three-by-one matrices with zero in the third row. If we let
\[\text{u}=\left(\begin{array}{c}u_1\\u_2\\0\end{array}\right),\quad\text{v}=\left(\begin{array}{c}v_1\\v_2\\0\end{array}\right),\nonumber \]
then
\[\text{w}=a\text{u}+b\text{v}=\left(\begin{array}{c}au_1+bv_1\\au_2+bv_2\\0\end{array}\right)\nonumber \]
is evidently also a three-by-one matrix with zero in the third row. This subspace of \(\mathbb{R}^3\) is closed under scalar multiplication and vector addition and is therefore a vector space. Another example of a vector subspace of \(\mathbb{R}^3\) would be the set of all three-by-one matrices where the first row is equal to the third row.
Of course, not all subsets of \(\mathbb{R}^3\) form a vector space. A simple example would be the set of all three-by-one matrices where the row elements sum to one. If, say, \(\text{u} =\left(\begin{array}{ccc}1&0&0\end{array}\right)^{\text{T}}\), then \(a\text{u}\) is a vector whose rows sum to \(a\), which can be different than one.
The zero vector must be a member of every vector space. If \(\text{u}\) is in the vector space, then so is \(0\text{u}\) which is just the zero vector. Another argument would be that if \(\text{u}\) is in the vector space, then so is \((−1)\text{u} = −\text{u}\), and \(\text{u} − \text{u}\) is again equal to the zero vector.
The concept of vector spaces is more general than a set of column matrices. Here are some examples where the vectors are functions.
Consider vectors consisting of all real polynomials in \(x\) of degree less than or equal to \(n\). Show that this set of vectors (together with the set of real numbers) form a vector space.
Solution
Consider the polynomials of degree less than or equal to \(n\) given by
\[p(x)=a_0+a_1x+a_2x^2+\cdots +a_nx^n,\quad q(x)=b_0+b_1x+b_2x^2+\cdots +b_nx^n,\nonumber \]
where \(a_0,\: a_1,\cdots , a_n\) and \(b_0,\: b_1,\cdots , b_n\) are real numbers. Clearly, multiplying these polynmials by real numbers still results in a polynomial of degree less than or equal to \(n\). Adding these polynomials results in
\[p(x)+q(x)=(a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2+\cdots +(a_n+b_n)x^n,\nonumber \]
which is another polynomial of degree less than or equal to \(n\). Since this set of polynomials is closed under scalar multiplication and vector addition, it forms a vector space. This vector space is designated as \(\mathbb{P}_n\).
Consider a function \(y = y(x)\) and the differential equation \(d^3y/dx^3 = 0\). Find the vector space associated with the general solution of this differential equation.
Solution
From Calculus, we know that the function whose third derivative is zero is a polynomial of degree less than or equal to two. That is, the general solution to the differential equation is
\[y(x)=a_0+a_1x+a_2x^2,\nonumber \]
which is just all possible vectors in the vector space \(\mathbb{P}_2\).
Consider a function \(y = y(x)\) and the differential equation \(d^2y/dx^2 + y = 0\). Find the vector space associated with the general solution of this differential equation.
Solution
Again from Calculus, we know that the trigonometric functions \(\cos x\) and \(\sin x\) have second derivatives that are the negative of themselves. The general solution to the differential equation consists of all vectors of the for
\[y(x)=a\cos x+b\sin x,\nonumber \]
which is just all possible vectors in the vector space consisting of a linear combination of \(\cos x\) and \(\sin x\).