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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.

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  • $\begingroup$ Perhaps it'll be better if someone simply clarifies the proof I posted up? A detailed walk-through of each step would be nice. :) $\endgroup$ – The Pointer Aug 23 '16 at 6:45
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You can consideradas the next explanation also for the fact that the row dimensión of a Matrix equals the column dimensión 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 ser 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 $A^T$ id 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.

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  • $\begingroup$ Thanks but I don't quite understand this explanation either. I'm not familiar with terms such as 'singular value decomposition'. $\endgroup$ – The Pointer Aug 22 '16 at 21:27
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    $\begingroup$ What the proof is basically saying is the next thing: it first takes the columns of the matrix $A$ and sees that there are k linearly independent columns, implying that all the other can be written as a linear combination of those k vectors. Then the columns of $A$ are expressed as the matricial product $BC$, where B is just the matrix whose columns are the $k-basis$ of the columns and $C$ is the matrix containing the coefficients necessary to create the columns of $A$ from the $k-basis$ vector. Then the reasoning is based that, as a matricial product can be seen as many diferent... $\endgroup$ – Josu Etxezarreta Martinez Aug 23 '16 at 8:11
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    $\begingroup$ ways for doing (see outer and inner product en.wikipedia.org/wiki/Matrix_multiplication), then it can be seen that the rows of A are spanned by the rows of your recently generated matrix C with the coefficients given by B. This implies that, as C is kxn, the dimension of the row space is as maximum k, implying that it has to be less or equal than the column space dimension. Finally, if you apply all of this reasoning to the $A^T$, you get the same result as before, but as the rows of A are the columns of $A^T$, you can get the last inequality... $\endgroup$ – Josu Etxezarreta Martinez Aug 23 '16 at 8:15
  • $\begingroup$ As both of the inequalities are true, that implies that both of dimensions have to be the same. Hope that this helps! $\endgroup$ – Josu Etxezarreta Martinez Aug 23 '16 at 8:16
  • $\begingroup$ Got it! Thank you. :) $\endgroup$ – The Pointer Aug 23 '16 at 17:08
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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 $$ \dim (\mathrm{colsp}\ A) = \dim (\mathrm{rowsp}\ A)\,. \quad \square $$

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