I have the following theorem:

Theorem 3.12. Let A be an m n matrix. Then the dimension of its row space is equal to the dimension of its column space.

And the following proof is given:

Proof. Suppose that $\lbrace v_1,v_2,\dots,v_k\rbrace$ is a basis for the column space of $A$. Then each column of $A$ can be expressed as a linear combination of these vectors; suppose that the $i$-th column $c_i$ is given by $$c_i = \gamma_{1i}v_1+\gamma_{2i}v_2+\dots+\gamma_{ki}v_k$$ Now form two matrices as follows: $B$ is an $m\times k$ matrix whose columns are the basis vectors $v_i$, while $C=(\gamma_{ij})$ is a $k\times n$ matrix whose $i$-th column contains the coefficients $\gamma_{1i},\gamma_{2i,}\dots,\gamma_{ki}$. It then follows$^7$ that $A=BC$.

However, we can also view the product $A= BC$ as expressing the rows of $A$ as a linear combination of the rows of $C$ with the $i$-th row of $B$ giving the coefficients for the linear combination that determines the $i$-th row of $A$. Therefore, the rows of $C$ are a spanning set for the row space of $A$, and so the dimension of the row space of $A$ is at most $k$. We conclude that: $$\dim(\operatorname{rowsp}(A))\leq\dim(\operatorname{colsp}(A))$$ Applying the same argument to $A^t$, we conclude that:$$\dim(\operatorname{colsp}(A))\leq\dim(\operatorname{rowsp}(A))$$and hence these values are equal

However, I am finding this proof impossible to follow and understand. Can someone please offer an alternative proof or explain what this proof is saying?

Thank you.

4 Answers

You can consider it as the next explanation also for the fact that the row dimension of a Matrix equals the column dimension of a matrix. For that I will use what it's called the rank of a Matrix.

The rank $r$ of a Matrix can be defines as the number of non-zero singular values of the Matrix, So applying the singular value decomposition of the matrix, we get $A=U\Sigma V^T$. This implies that the range $dim(R(A))=r$, as the range of $A$ is spanned by the first $r$ columns of $U$. We know that the range of $A$ is defined as the subspace spanned by the columns of $A$, so the dimension of it will be $r$.

If we take the transpose of the Matrix and compute it's SVD, we see that $A^T=V\Sigma^T U^T$, and as the Sigma Matrix remains the same number of non-zero elements as the one for $A$, the rank of this Matrix will still be $r$. So as done for $A$, the dimension for the range of $A^T$ is equal to $r$ too, but as the range of $A^T$ is the row space of $A$, we conclude that the dimension for both spaces must be the same and equal to the range of the Matrix $A$.

Here is a more explicit step-by-step version of the proof quoted in the question.

Proof

Let $A\in\mathcal{M}_{m\times n}(\mathbb{K})$, where $\mathbb{K}$ is any field. Let $\{\boldsymbol{u^1},\ldots, \boldsymbol{u^k} \}$ be a basis of the column space of $A$, where $k\in\{1,\ldots,n\}$. $\boldsymbol{u^i}$ is a vector in $\mathbb{K}^m$ (the vector space of $m$-tuples with entries in $\mathbb{K}$) for all $i\in\{1,\ldots,k\}$. Thus each column in $A$ can be expressed as a linear combination of $\boldsymbol{u^1},\ldots,\boldsymbol{u^k}$. That is, for every $j\in\{1,\ldots,n\}$ there exist unique coefficients $\lambda_{1j},\ldots,\lambda_{nj}\in\mathbb{K}$ such that $$ \forall j\in\{1,\ldots,n\}\qquad \boldsymbol{v^j} = \sum_{\ell=1}^{k} \lambda_{\ell j} \boldsymbol{u^\ell}\,, $$ where $\boldsymbol{v^j}$ denotes the $j$-th column in $A$.

Let $B\in\mathcal{M}_{m\times k}(\mathbb{K})$ be the matrix with such vectors as columns: $$ B = \begin{pmatrix} \vert & & \vert \\ \boldsymbol{u^1} & \cdots & \boldsymbol{u^k} \\ \vert & & \vert \end{pmatrix}\,. $$ Let $[s]$ denote $\{1,\ldots,s\}$ for all $s\in\mathbb{N}$. Let $C\in\mathcal{M}_{k\times n}(\mathbb{K})$ be the matrix with the aforementioned coefficients: $$ C = (\lambda_{ij})_{(i,j)\in[k]\times[n]} = \begin{pmatrix} \lambda_{11} & \cdots & \lambda_{1n} \\ \vdots & \ddots & \vdots \\ \lambda_{k1} & \cdots & \lambda_{kn} \end{pmatrix}\,. $$

Now consider the matrix product $BC\in\mathcal{M}_{m\times n}(\mathbb{K})$. Let $(bc)_{ij}$, $b_{ij}$ and $c_{ij}$ denote the $(i,j)$-th element of $BC$, $B$ and $C$ respectively. By definition of matrix product, \begin{equation}\tag{1}\label{foo} (bc)_{ij} = \sum_{\ell=1}^{k} b_{i\ell}c_{\ell j} \qquad\forall (i,j)\in[m]\times[n]\,. \end{equation} Let us consider the $j$-th column of $BC$ for an arbitrary $j\in[n]$. Let $v^\ell_i$ denote the $i$-th component of $\boldsymbol{u^\ell}$ for all $\ell\in[k]$ and for all $i\in[m]$.

$$ \begin{multline*} \left((bc)_{ij}\right)_{i\in[m]} = \left( \sum_{\ell=1}^{k} b_{i\ell}c_{\ell j} \right)_{i\in[m]} = \begin{pmatrix} \sum_{\ell=1}^{k} b_{1\ell}c_{\ell j} \\ \vdots \\ \sum_{\ell=1}^{k} b_{m\ell}c_{\ell j} \end{pmatrix} = \\ % = \begin{pmatrix} \sum_{\ell=1}^{k} v^\ell_1\cdot\lambda_{\ell j} \\ \vdots \\ \sum_{\ell=1}^{k} v^\ell_m\cdot\lambda_{\ell j} \end{pmatrix} = \sum_{\ell=1}^k \lambda_{\ell j} % \begin{pmatrix} v^\ell_1 \\ \vdots \\ v^\ell_m \end{pmatrix} = \sum_{\ell=1}^{k} \lambda_{\ell j}\boldsymbol{u^\ell} = \boldsymbol{v^j}\,. \end{multline*} $$

Thus, the columns of $BC$ are the columns of $A$. Ergo, $A=BC$.

On the other hand, let us consider the $i$-th row of $A$, denoted by $\boldsymbol{r^i}$. That is, $$ \boldsymbol{r^i} = (a_{ij})_{j\in[n]} \qquad \forall i\in[m]\,. $$ Again, by the definition of matrix multiplication, $$ a_{ij} = \sum_{\ell=1}^{k} b_{i\ell}c_{\ell j} \qquad\forall (i,j)\in[m]\times[n] $$ (this is the same equation found in eq. \eqref{foo}). Thus,

$$ \begin{multline}\tag{2}\label{rows} \boldsymbol{r^i} = \left(\sum_{\ell=1}^{k} b_{i\ell}c_{\ell j} \right)_{j\in[n]} = \begin{pmatrix} \sum_{\ell=1}^{k} b_{i\ell}c_{\ell 1} & \cdots & \sum_{\ell=1}^{k} b_{i\ell}c_{\ell n} \end{pmatrix} = \\ = \begin{pmatrix} \sum_{\ell=1}^{k} v^\ell_i\cdot\lambda_{\ell 1} & \cdots & \sum_{\ell=1}^{k} v^\ell_i\cdot\lambda_{\ell n}\,. \end{pmatrix} \end{multline} $$

Now, let $\boldsymbol{\Lambda^\ell}$ be the $\ell$-th row of $C$ for all $\ell\in[k]$, as a row vector:

$$ \begin{equation*} \boldsymbol{\Lambda^\ell} = \begin{pmatrix} \lambda_{\ell 1} & \cdots & \lambda_{\ell n}\,. \end{pmatrix} \end{equation*} $$

Thus, with the same notation as before, ${\Lambda^\ell_i} = \lambda_{\ell i}$ for all $i\in[n]$. Also, let $\mu_{i \ell}$ denote $v^\ell_i$ for all $i\in[m]$ and for all $\ell\in[k]$ — this is merely a change of notation to remark the fact that $v^\ell_i$ can be seen as "coefficients". Thus, continuing to develop equation \eqref{rows}, we get

$$ \begin{multline*} \boldsymbol{r^i} = \begin{pmatrix} \sum_{\ell=1}^{k} \mu_{i\ell}\cdot\Lambda^\ell_1 & \cdots & \sum_{\ell=1}^{k} \mu_{i\ell}\cdot\Lambda^\ell_n \end{pmatrix} = \\ = \sum_{\ell=1}^k \mu_{i\ell} \begin{pmatrix} \Lambda^\ell_1 & \cdots & \Lambda^\ell_n \end{pmatrix} = \sum_{\ell=1}^k \mu_{i\ell} \boldsymbol{\Lambda^\ell}\,. \end{multline*} $$

Therefore, the rows of $A$ (i.e., $\boldsymbol{r^i}$) are linear combinations of the rows of $C$ (i.e., $\boldsymbol{\Lambda^\ell}$). Thus, we necessarily have $$ \mathrm{rowsp}\ A \subseteq \mathrm{rowsp}\ C \ \implies\ \dim (\mathrm{rowsp}\ A) \le \dim (\mathrm{rowsp}\ C)\,. $$ Since $C$ has $k$ rows, its row space can have at most dimension $k$, which is the dimension of $\mathrm{colsp}\ A$ (by hypothesis): $$ \dim(\mathrm{rowsp}\ C) \le k = \dim(\mathrm{colsp}\ A)\,. $$ Combining both inequalities, we have $$ \dim (\mathrm{rowsp}\ A) \le \dim (\mathrm{colsp}\ A)\,. $$

Applying this whole argument again on $A^\mathrm{t}$, $$ \dim (\mathrm{rowsp}\ A^\mathrm{t}) \le \dim (\mathrm{colsp}\ A^\mathrm{t}) \iff \dim (\mathrm{colsp}\ A) \le \dim (\mathrm{rowsp}\ A)\,. $$ Since we have both $\dim (\mathrm{rowsp}\ A) \le \dim (\mathrm{colsp}\ A)$ and $\dim (\mathrm{colsp}\ A) \le \dim (\mathrm{rowsp}\ A)$, we conclude that $$ \dim (\mathrm{colsp}\ A) = \dim (\mathrm{rowsp}\ A)\,. \quad \square $$

The long proof by Anakhand, even though perfectly correct, is harder to understand than the proof given in the original question. This is because it gets to the nitty-gritty details of matrix multiplication. Matrix multiplication can be understood in several different ways. One of them is the fact that if a matrix is right-multiplied by a column vector, then the output is a linear combination of columns of the matrix where the coefficients are the vector's coordinates (this is crucial to understanding the proof in the question and is a simple observation). So, if one multiplies 2 matrices, the output is a matrix where each column of the output matrix is a linear combination of the columns of the first matrix and the coefficients are the coordinates of the corresponding vector in the second matrix. On the other hand, if you left-multiply a matrix by a row vector, then the output is a row vector which is a linear combination of rows in the matrix. Similarly when you multiply 2 matrices. If one understands this, the proof in the question becomes much clearer than the proof that goes into the nitty-gritty details of matrix multiplication.

Row reduction eliminates the redundant rows in a matrix, i.e. the rows that are linear combinations of other rows. Row operations also do not change the row space because taking nonzero linear combinations of a set of vectors does not change their span.

It is impossible not to eliminate a row that is a linear combination of the others when reducing to reduced echelon form, and so the RREF of the matrix will give you the number of linearly independent rows, i.e. the dimension of the row space. Let p be the number of rows of zeroes at the bottom of the matrix. Then the dimension of the row space is m - p.

Now consider the effect of this reduction on columns. Row operations preserve linear dependence relations among the columns of A so that the same linear dependence relation holds for columns of the RREF of A as for A. If you have an m x n matrix with p rows of zeros at the bottom, then you also have m - p pivots. If this were not the case, then one of the rows above those p rows would be row reducible to a row of all zeros, which is what happens to any row that does not have a pivot.

Since the number of pivots is equal to the number of linearly independent vectors in the RREF, and this is also equal to the number of linearly independent vectors in the original matrix, these columns form a basis for Col(A). The dimension of the column space is the number of basis vectors, and so we have shown that the two are equal.