Vector spaces and linear maps – university Year 1 maths lesson to download as PDF

Example 1

Which of \(\{x+y+z=0\}\), \(\{xy=0\}\) and \(\{x\ge0\}\) are subspaces?

The plane \(\{(x,y,z):x+y+z=0\}\) is a subspace of \(\mathbb R^3\): if two vectors have coordinates summing to 0, so do their sum and any multiple. \(\{(x,y):xy=0\}\) is not, since \((1,0)+(0,1)=(1,1)\) is not in it. \(\{(x,y):x\ge0\}\) fails (iii) with \(c=-1\).

Example 2

Do \((1,2,3)\), \((0,1,4)\), \((5,6,0)\) form a basis of \(\mathbb R^3\)? What about \((1,2,3)\), \((4,5,6)\), \((7,8,9)\)?

The first set has determinant 1, so it is a basis. In the second, \(\mathbf v_1-2\mathbf v_2+\mathbf v_3=\mathbf 0\), so the vectors are dependent.

Example 3: coordinates

Find the coordinates of \((2,3,4)\) in the basis \((1,1,0),(0,1,1),(1,0,1)\).

Solve \(a+c=2\), \(a+b=3\), \(b+c=4\). The coordinates are \((\tfrac12,\tfrac52,\tfrac32)\); they are unique for each vector.

Example 4

Find the matrix of \(T(x,y)=(x+y,2x-y)\), and of differentiation \(D:P_2\to P_2\).

\(T\) has matrix \(\begin{pmatrix}1&1\\2&-1\end{pmatrix}\). In the basis \(1,x,x^2\), \(D\) has matrix \(\begin{pmatrix}0&1&0\\0&0&2\\0&0&0\end{pmatrix}\).

Example 5

For \(T(x,y,z)=(x+y,y+z)\) find the kernel, image, rank and nullity.

The kernel is spanned by \((1,-1,1)\) (nullity 1) and the image is all of \(\mathbb R^2\) (rank 2). Check: \(3=2+1\).

Worked example 6: Testing for a basis with a determinant

Do \((1,0,2)\), \((2,1,0)\) and \((0,3,1)\) form a basis of \(\mathbb R^3\)?

Three vectors in \(\mathbb R^3\) are a basis exactly when they are linearly independent, which happens exactly when the determinant of the matrix of their coordinates is non-zero. Here \(\begin{vmatrix}1&0&2\\2&1&0\\0&3&1\end{vmatrix}=1(1-0)-0+2(6-0)=13\ne0\), so they do form a basis. If the determinant had been zero we would look for a dependence relation between them.

Worked example 7: The span of two vectors

Describe the span of \(\mathbf a=(1,2,3)\) and \(\mathbf b=(0,1,1)\) in \(\mathbb R^3\) and decide whether \((2,5,7)\) lies in it.

The vectors are not multiples of each other, so their span is a plane through the origin. A vector is in the span if \((2,5,7)=\lambda\mathbf a+\mu\mathbf b\). The first coordinate gives \(\lambda=2\); the second then gives \(4+\mu=5\), so \(\mu=1\); the third coordinate gives \(6+1=7\), which agrees. So yes: \((2,5,7)=2\mathbf a+\mathbf b\). For \((2,5,8)\) the same method would fail at the third coordinate, so that vector is outside the plane.

Worked example 8: The kernel of a map between polynomial spaces

Let \(T:P_2\to P_2\) be \(T(p)=p''+p\). Find its kernel and image.

Take \(p=a+bx+cx^2\). Then \(p''=2c\), so \(T(p)=(a+2c)+bx+cx^2\). Setting this equal to zero forces \(b=0\), \(c=0\) and then \(a=0\), so the kernel is \(\{0\}\). By rank-nullity the rank is \(3\), so the image is all of \(P_2\): for any target polynomial one can solve for \(a,b,c\). A map with trivial kernel is injective, and here also surjective, so it is an isomorphism.

Worked example 9: Rotation and its matrix

Find the matrix of the rotation through \(90^\circ\) anticlockwise in the plane, and its kernel and image.

The images of the basis vectors are \((1,0)\mapsto(0,1)\) and \((0,1)\mapsto(-1,0)\). These are the columns of the matrix, so it is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\). Its determinant is \(1\) and no non-zero vector is sent to zero, so the kernel is \(\{\mathbf 0\}\) and the image is all of \(\mathbb R^2\). The map is bijective, and its fourth power is the identity, because four quarter-turns make a full turn.

Worked example 10: Rank-nullity for a projection

The map \(P(x,y,z)=(x,y,0)\) projects space onto the horizontal plane. Find its rank and nullity.

The image is the \(xy\)-plane, which has dimension 2, so the rank is 2. The kernel is the \(z\)-axis, since \(P(0,0,z)=\mathbf 0\), which has dimension 1. Check: \(2+1=3=\dim\mathbb R^3\). Notice that applying \(P\) twice gives the same result as applying it once, which is the algebraic signature of a projection.