Linear algebra is the language of data, graphics, networks, and differential equations. Its central idea is simple: many different objects (arrows, polynomials, matrices, functions) obey the same rules of addition and scaling, so one theory describes them all. In this chapter you will learn to recognize these spaces, measure their size, and understand the maps between them.
1. Vector spaces and subspaces
A vector space over \(\mathbb{R}\) is a set \(V\) with two operations, addition \(u+v\) and scalar multiplication \(cu\), that never leave \(V\) and satisfy the usual rules: addition is commutative and associative, there is a zero vector \(\mathbf{0}\), every \(v\) has an opposite \(-v\), and scalars distribute over sums and over each other, with \(1\cdot v = v\).
Typical examples are \(\mathbb{R}^n\), the set \(P_n\) of polynomials of degree at most \(n\), the set \(M_{m\times n}\) of \(m\times n\) matrices, and the set of continuous functions on an interval.
A subset \(W\) of a vector space \(V\) is a subspace when \(W\) is itself a vector space under the operations of \(V\).
A subset \(W \subseteq V\) is a subspace if and only if (i) \(\mathbf{0}\in W\), (ii) \(u+v\in W\) whenever \(u,v\in W\), and (iii) \(cu\in W\) whenever \(u\in W\) and \(c\in\mathbb{R}\).
Let \(W=\{(x,y,z)\in\mathbb{R}^3 : x+2y-z=0\}\). First, \(0+0-0=0\), so \(\mathbf{0}\in W\). If \((x_1,y_1,z_1)\) and \((x_2,y_2,z_2)\) are in \(W\), adding the two equations gives \((x_1+x_2)+2(y_1+y_2)-(z_1+z_2)=0\), so the sum is in \(W\). Multiplying the equation by \(c\) shows that \(cu\in W\). Hence \(W\) is a subspace. By contrast, the set \(x+2y-z=1\) is not a subspace because it does not contain \(\mathbf{0}\).
Violet: the line \(y=x/2\), a subspace of \(\mathbb{R}^2\). Orange: the line \(y=x+1\), which misses the origin and is not a subspace.
Checking closure is not enough: always check that the zero vector belongs to the set. Also, the union of two subspaces is usually not a subspace: the \(x\)-axis together with the \(y\)-axis contains \((1,0)\) and \((0,1)\) but not their sum \((1,1)\).
2. Span and linear independence
The span of vectors \(v_1,\dots,v_k\) is the set of all linear combinations \(c_1v_1+\dots+c_kv_k\). It is always a subspace. The vectors are linearly independent if the only solution of \(c_1v_1+\dots+c_kv_k=\mathbf{0}\) is \(c_1=\dots=c_k=0\); otherwise they are linearly dependent.
- Write the vectors as the columns of a matrix \(A\).
- Row reduce \(A\) to echelon form.
- The vectors are independent exactly when every column contains a pivot.
Take \(v_1=(1,2,3)\), \(v_2=(0,1,1)\), \(v_3=(1,3,4)\). Notice that \(v_1+v_2=(1,3,4)=v_3\), so \(v_1+v_2-v_3=\mathbf{0}\) is a nontrivial relation. The family is dependent, and \(v_3\) lies in the span of \(v_1\) and \(v_2\). Removing \(v_3\) leaves two independent vectors, because neither is a multiple of the other.
Any family containing \(\mathbf{0}\) is dependent, and two vectors are dependent exactly when one is a multiple of the other. Pairwise checks are not enough for three or more vectors: in Example 2 no two vectors are multiples of each other, yet the three together are dependent.
3. Basis and dimension
A basis of \(V\) is a family that is linearly independent and spans \(V\). Every vector of \(V\) can then be written in exactly one way as a combination of the basis vectors.
All bases of a finite-dimensional space \(V\) contain the same number of vectors; this number is the dimension \(\dim V\). If \(\dim V=n\), then any \(n\) independent vectors form a basis, any \(n\) spanning vectors form a basis, and any family with more than \(n\) vectors is dependent.
Familiar dimensions: \(\dim\mathbb{R}^n=n\), \(\dim P_n=n+1\) (basis \(1,x,\dots,x^n\)), and \(\dim M_{m\times n}=mn\).
In Example 1, the condition \(x+2y-z=0\) gives \(z=x+2y\), so \((x,y,z)=x(1,0,1)+y(0,1,2)\). The vectors \((1,0,1)\) and \((0,1,2)\) are independent and span \(W\), so they form a basis and \(\dim W=2\): the plane has two dimensions, as it should. The free variables \(x\) and \(y\) count the dimension.
4. Null space and column space
For an \(m\times n\) matrix \(A\), the null space is \(\mathrm{Nul}(A)=\{x\in\mathbb{R}^n : Ax=\mathbf{0}\}\), a subspace of \(\mathbb{R}^n\). The column space \(\mathrm{Col}(A)\) is the span of the columns of \(A\), a subspace of \(\mathbb{R}^m\). The rank of \(A\) is \(\dim\mathrm{Col}(A)\), and the nullity is \(\dim\mathrm{Nul}(A)\).
- Row reduce \(A\) to reduced echelon form.
- For \(\mathrm{Nul}(A)\): solve \(Ax=\mathbf{0}\), give each free variable a parameter, and read off one basis vector per free variable.
- For \(\mathrm{Col}(A)\): take the columns of the original matrix \(A\) that sit in the pivot positions.
Let \(A=\begin{pmatrix}1&2&1&3\\2&4&3&8\\1&2&0&1\end{pmatrix}\). Row reduction gives \(\begin{pmatrix}1&2&0&1\\0&0&1&2\\0&0&0&0\end{pmatrix}\). The pivots are in columns 1 and 3, so \(\mathrm{Col}(A)\) has basis \((1,2,1)\) and \((1,3,0)\), and the rank is 2. The free variables are \(x_2=s\) and \(x_4=t\); then \(x_1=-2s-t\) and \(x_3=-2t\), so \(\mathrm{Nul}(A)\) has basis \((-2,1,0,0)\) and \((-1,0,-2,1)\), and the nullity is 2.
5. The rank-nullity theorem
For an \(m\times n\) matrix \(A\), \[\operatorname{rank}(A)+\operatorname{nullity}(A)=n,\] the number of columns. Equivalently, for a linear map \(T:V\to W\), \(\dim\ker T+\dim\operatorname{range}T=\dim V\).
The theorem says that the dimensions of the domain are shared between what is crushed to zero (the null space) and what survives (the column space). In Example 4, \(2+2=4\), the number of columns of \(A\).
Zyro, the alien explorer, counts the “lost directions” first: with \(n\) columns and \(r\) pivots, exactly \(n-r\) free variables remain, and that is the dimension of the null space. Count free variables and you have the nullity for free!
The \(n\) in the theorem is the number of columns (the dimension of the domain), not the number of rows. A \(3\times 5\) matrix of rank 2 has nullity \(5-2=3\).
6. Linear transformations
A map \(T:V\to W\) is linear if \(T(u+v)=T(u)+T(v)\) and \(T(cu)=cT(u)\) for all \(u,v\in V\) and \(c\in\mathbb{R}\). In particular \(T(\mathbf{0})=\mathbf{0}\). Its kernel is \(\ker T=\{v : T(v)=\mathbf{0}\}\) and its range is \(\{T(v): v\in V\}\).
Every linear map \(T:\mathbb{R}^n\to\mathbb{R}^m\) is multiplication by a unique matrix \(A\), whose \(j\)-th column is \(T(e_j)\). Then \(\ker T=\mathrm{Nul}(A)\) and the range is \(\mathrm{Col}(A)\). The map is one-to-one exactly when \(\ker T=\{\mathbf{0}\}\), and onto exactly when the rank equals \(m\).
Let \(T(x,y)=(x+y,\,2x-y,\,3y)\). Then \(T(e_1)=T(1,0)=(1,2,0)\) and \(T(e_2)=T(0,1)=(1,-1,3)\), so \(A=\begin{pmatrix}1&1\\2&-1\\0&3\end{pmatrix}\). The two columns are independent, so the rank is 2 and, by rank-nullity, the nullity is \(2-2=0\): \(T\) is one-to-one. It is not onto \(\mathbb{R}^3\), since the rank is 2, not 3. Note that \(T(0,0)=(0,0,0)\), as every linear map requires; the map \(S(x,y)=(x+1,y)\) is therefore not linear.
The matrix with columns \((2,0)\) and \((1,1)\) maps the unit square to a parallelogram. Its area is \(|\det|=2\), so areas are doubled.
7. Change of basis
A vector can be described by different lists of numbers, depending on the basis you choose. If \(B=\{b_1,\dots,b_n\}\) is a basis of \(\mathbb{R}^n\), the coordinates of \(v\) in \(B\) are the numbers \(c_1,\dots,c_n\) with \(v=c_1b_1+\dots+c_nb_n\); we write \([v]_B=(c_1,\dots,c_n)\).
Let \(P\) be the matrix whose columns are \(b_1,\dots,b_n\). Then \(v=P[v]_B\), that is \([v]_B=P^{-1}v\). If \(T\) has standard matrix \(A\), its matrix in the basis \(B\) is \([T]_B=P^{-1}AP\).
Let \(b_1=(2,1)\), \(b_2=(1,1)\), so \(P=\begin{pmatrix}2&1\\1&1\end{pmatrix}\) with \(\det P=1\) and \(P^{-1}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}\). For \(v=(5,3)\), \([v]_B=P^{-1}v=(5-3,\,-5+6)=(2,1)\). Check: \(2(2,1)+1(1,1)=(5,3)\). Now let \(A=\begin{pmatrix}3&1\\0&2\end{pmatrix}\). Then \(AP=\begin{pmatrix}7&4\\2&2\end{pmatrix}\) and \([T]_B=P^{-1}AP=\begin{pmatrix}5&2\\-3&0\end{pmatrix}\). The trace (5) and determinant (6) are unchanged, as they must be.
Going twice along \(b_1\) and once along \(b_2\) reaches \(v=(5,3)\).
- Build \(P\) with the basis vectors as columns.
- To go from standard to \(B\)-coordinates, solve \(Pc=v\) (or compute \(P^{-1}v\)).
- To go back, compute \(v=Pc\).
Key takeaways
- A subspace contains \(\mathbf{0}\) and is closed under addition and scalar multiplication.
- Vectors are independent when the only combination giving \(\mathbf{0}\) is the trivial one; a basis is an independent spanning family, and \(\dim V\) is its size.
- \(\mathrm{Nul}(A)\) lives in the domain and \(\mathrm{Col}(A)\) lives in the codomain; pivot columns of \(A\) give a basis of \(\mathrm{Col}(A)\).
- Rank-nullity: \(\operatorname{rank}+\operatorname{nullity}=\) number of columns.
- A linear map is determined by the images of a basis; its matrix has columns \(T(e_j)\); it is one-to-one iff its kernel is \(\{\mathbf{0}\}\).
- Change of basis: \([v]_B=P^{-1}v\) and \([T]_B=P^{-1}AP\).
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