I am confused about this matrix. We know Row rank = column rank = determinant rank for a matrix and proof is known to all. See the following matrix

$$A = \left[\begin{array} &t &t^2\\ 0 &0 \end{array}\right]$$

Here $\det(A) = 0 $. Row rank = 1. Column rank =2 determinant rank =1.

Am I true? If not make a correction.

Here columns are independent but $\det(A) = 0$? Is it contradictory to the result? Please discuss.

This matrix is used as a counterexample in the book of "Theory of Ordinary Differential Equation" by "Coddinton & Livenson" to show that $\det(A)=0$ but columns are independent.

Please discuss. Thank you.

  • 2
    $\begingroup$ The columns aren't independent, $t\begin{pmatrix}t\\0\end{pmatrix} - 1\begin{pmatrix}t^2\\0\end{pmatrix} = 0$. $\endgroup$ Aug 7, 2013 at 12:37
  • $\begingroup$ @Samprity: Where in the C&L book is that example? $\endgroup$
    – Amzoti
    Aug 7, 2013 at 12:49
  • $\begingroup$ The columns are independent over $\mathbb{R}$. Just like $\{ \sin(t) \}$ is linearly independent over $\mathbb{R}$ even though $\sin(\pi)=0$. I think the clear context here is function space and differential equations, probably around the discussion of the Wronskian... $\endgroup$ Aug 7, 2013 at 13:40
  • 1
    $\begingroup$ Is it supposed to be Levinson instead of Livenson? $\endgroup$ Aug 7, 2013 at 13:41
  • $\begingroup$ Theory of Ordinary Differential Equation by E. A. Coddington, N. Levinson Tata McGraw-Hill Publishing Company Limited 32nd reprint 2009 Chapter 3 Linear Differential Equation Page 70. $\endgroup$
    – Supriyo
    Aug 7, 2013 at 14:00

3 Answers 3


The issue here is what is the field over which we are working. Also, you are assuming that finite dimensional results apply to infinite dimensional linear algebra, a dangerous game.

Yes, $\{ (t,0), (t^2,0) \}$ are linearly independent as vectors over $\mathbb{R}$ in infinite dimensional function space. The proof is simple: $$ c_1(t,0)+c_2(t^2,0)= 0 $$ for all $t \in \mathbb{R}$ implies $c_1=c_2=0$. To see this, differentiate once, $$ c_1(1,0)+c_2(2t,0)= 0 \qquad \star $$ and then again: $$ c_1(0,0)+c_2(2,0)= 0 \qquad \star^2 $$ and evaluate $\star$ at $t=0$ to obtain $c_1=0$ and $\star^2$ at $t=0$ to obtain $c_2=0$. There are many other ways, I'm just fond of this argument.

Ok, so we have established the linear independence of the columns. Moreover, you are certainly correct that $\{ [t,t^2], [0,0] \}$ is linearly dependent, and of course has rank $1$.

Finally, yes $det\left[ \begin{array}{cc} t & t^2 \\ 0 & 0 \end{array} \right]=0$. So, what does this mean? This example shows that zero determinant does not prove linear dependence on function space. We can have a full-rank set of vectors of functions for which the determinant is zero and yet the set of functions is linearly independent.

Does this contradict linear algebra? No. In your earlier studies, you should recall we proved things for finite dimensional spaces over $\mathbb{R}$. This is not that context.

Interestingly, the story for linearly independent functions is different. If $\{ v_1, \dots ,v_n \}$ are linearly independent $\mathbb{R}^n$-valued functions on $\mathbb{R}$ (the domain is important in these questions, we must supply both formulas and domain) then it follows that $$ c_1v_1(t)+ \cdots + c_nv_n(t)=0 $$ for all $t$. Hence $[v_1(t)|\cdots | v_n(t)][c_1,\dots,c_n]^T=0$ has only the zero solution hence $[v_1(t)|\cdots | v_n(t)]$ is invertible for all $t$ and it follows that $det[v_1(t)|\cdots | v_n(t)] \neq 0$ for all $t$.

So, we find the curious situation that nonzero determinants are informative, but zero determinants are an invitation to further analysis. Obviously, there is great potential for confusion here.

  • 1
    $\begingroup$ Unless something unusual is going on (no sign of that in the question), entries of a matrix are scalars, not vectors. The fact that $t$ occurs as matrix entry means it is a scalar, and the comment by Daniel Fischer to the question always exhibits a bona fide linear dependence of the columns (the first column $\binom t0$ stands for $te_1+0e_2$ with $(e_1,e_2)$ the basis on which the matrix is expressed, so $t$ is a coefficient here; if it can be once, it can be so once again). This applies if $t\in\Bbb R$ for a real vector space, or $t$ is a function in a vector space over a function field. $\endgroup$ Aug 7, 2013 at 13:59
  • $\begingroup$ @James S. Cook Thank you very much for your clarification. This answer has given me some new ideas related to linear dependence and independence in various dimentions. $\endgroup$
    – Supriyo
    Aug 7, 2013 at 14:13
  • $\begingroup$ @MarcvanLeeuwen I totally agree that for fixed $t\in \mathbb{R}$ the column vectors $(t,0)$ and $(t^2,0)$ are linearly dependent, indeed scalar multiples. However, if $t$ is a variable and in fact what is considered is the linear independence of $S=\{f_1,f_2 \}$ where $f_1(t)=(t,0)$ and $f_2(t)=(t^2,0)$ then $S$ is LI over $\mathbb{R}$. Furthermore, since this question is from DEqns text this is almost certainly the intent of it being a "counterexample". We could say $\{ t,t^2 \}$ is LI just the same... this being short-hand for functions $g_1,g_2$ with the formulas $g_1(t)=t$, $g_2(t)=t^2$. $\endgroup$ Aug 7, 2013 at 16:00
  • $\begingroup$ @JamesS.Cook: I'm sorry to have to say this bluntly, but you are just plain wrong. Matrices and vector spaces are over some field (or ring, but using dimension requires a field). It has not been specified here and could be $\def\R{\Bbb R}\R$, $\R(t)$ (rational functions in $t$ over $\R$) or another function field. Anyway using a $2\times2$ matrix implies a space of dimension$~2$ over that field. Also $t$ must be in the field as it is a matrix entry, so it can be used as scalar. When considering functions as $\R$-vector space, one cannot use a $2\times2$ matrix with entries $t$, $t^2$. $\endgroup$ Aug 7, 2013 at 16:18
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    $\begingroup$ @MarcvanLeeuwen Again Marc, look at the OP's question. He is quoting a problem from a DIFFERENTIAL EQUATIONS text. The text claims that this is gives a counterexample showing LINEARLY INDEPENDENT columns but a zero determinant. Since the book says the columns are linearly independent, it must be working in the context of a function space not scalars. $\endgroup$
    – Bill Cook
    Aug 9, 2013 at 14:00

Column rank is the maximum number of columns which a linearly independent. Row rank is similarly defined. In your example, column rank is also $1$ because $(t, 0)$ and $(t^2, 0)$ are not linearly independent as stated in the comments. So it does not contradict the fact that column rank $=$ row rank. I'm not sure what is a determinant rank.

  • $\begingroup$ what is your context? in which vector space are you judging the linear independence of the set of vectors? This matters since if $t$ is a function then it is not a scalar. Unless, you intend to work over the module with base ring of functions... in any event, these issues must be discussed. $\endgroup$ Aug 7, 2013 at 13:42

There is some confusion in the question. This is partly due because the context is not well described, and partly due to ambiguity between writing a matrix with functions (of$~t\in\Bbb R)$ as entries or a function of$~t$ whose values are matrices with real entries. However the essence remains the same whatever point of view is adapted: the matrix $A$ of the question has linearly dependent rows, linearly dependent columns, and zero determinant.

Characterisation of linear dependency of rows or columns of a matrix by vanishing of the determinant is valid for square matrices over any field or even integral domain$~R$. Linear dependence of elements of $R^n$ (or any module over$~R$) means that some non-trivial $R$-linear combination of them is$~0$, just as in the vector space case. (One must exclude rings with zero divisors here; even giving a reasonable definition of linear dependence is not obvious over rings with zero divisors.)

Proposition. Let $A$ be a $n\times n$ matrix over an integral domain$~R$. The following are equivalent.

  1. The rows of $A$ are $R$-linearly dependent.
  2. The columns of $A$ are $R$-linearly dependent.
  3. $\det A=0\in R$.

Proof. The implications $1.\Rightarrow3.$ and $2.\Rightarrow3.$ follow from the multi-linear and alternating character of the determinant with respect to rows respectively columns: for $1.\Rightarrow3.$, if the rows $r_i$ of $A$ are linearly dependent then some nonzero multiple $cr_i$ of a row is equal to a linear combination of the remaining rows; applying linearity of the determinant it follows that $c\det A$ is a linear combination of determinants in all of which two rows are equal, so it is zero, and thus $\det A=0$ because $c\neq0$. For the reverse implications in the contrapositive form $\lnot1.\Rightarrow\lnot3.$ and $\lnot2.\Rightarrow\lnot3.$, it is easiest to use the field of fractions $F=\operatorname{Frac}(R)$ of$~R$: from the linear independence it follows that the row (resp. column) space of $A$ over $F$ has dimension$~n$, so is all of$~F^n$, and in particular contains the standard basis of$~F^n$; taking $B\in \operatorname{Mat}_n(F)$ to be the matrix whose rows (resp. columns) hold the coefficients of expression of the standard basis vectors in terms of the rows (columns) of$~A$ one then has $BA=I_n$ (resp. $AB=I_n$) and taking determinants shows that $\det A$ is invertible in$~F$, so nonzero in$~R$. QED

(For the reverse implication which is most pertinent to this question, there also exists a more constructive argument that avoids using the field of fractions, or even that $R$ is a domain, and which by clever use of cofactors deduces from $\det A=0$ an explicit nontrivial relation between the rows or columns of$~A$.)

I've insisted a bit to show that "the columns (or rows) of a singular matrix are always linearly dependent" really should be considered a general fact, not something that may or may not hold according to the context. One may however force it to fail artificially by qualifying "linear dependent" by a field (or ring) that is too small to contain the entries of the matrix; for instance the columns of the singular real matrix $$ M=\begin{pmatrix}1&\sqrt 2\\ \sqrt2&2\end{pmatrix} $$ are linearly independent over the rational numbers. But while this is formally correct (viewing $\Bbb R^2$ as vector space over$~\Bbb Q$), it is a bit pointless to say so, since just viewing the real numbers as a vector space over the rational numbers is insufficient to even compute $\det M=0$ (and as soon as one enlarges $\Bbb Q$ as base field to $\Bbb Q(\sqrt2)$, the columns of$~M$ do become linearly dependent).

So to conclude the answer to the question as posed: for all reasonable ends saying that the columns of the matrix $$A=\begin{pmatrix}t&t^2\\ 0&0\end{pmatrix}$$ are linearly independent, or that its column rank is$~2$ is just a plain mistake, whatever $t$ is meant to stand for (the matrix obviously has zero determinant and linearly dependent rows, and its columns are (necessarily) also linearly dependent with for instance a dependence relation with coefficients $t,-1$; saying that $t$ is forbidden as coefficient in a relation while it is allowed as matrix entry would be as silly as talking about rational independence of the columns of the non-rational matrix$~M$ above).

Although it is not very explicit in the question, I can somewhat guess where your confusion comes from. In the context of differential equations, you might want to reason about $\Bbb R$-linear independence of functions, for instance in order to show that they span the entire solution space (assuming that is a subspace, within some infinite dimensional space of functions where the equations are defined, which subspace has a known finite dimension). In an infinite dimensional context one cannot hope that linear independence is controlled by a single (finite) determinant as in the proposition above. One can however hope that the non-vanishing of a properly constructed determinant provides a sufficient condition for linear independence. Indeed if $v_1,\ldots,v_n$ are vectors in any $\Bbb R$-vector space$~V$, and if for some list $\alpha_1,\ldots,\alpha_n$ of linear forms $V\to\Bbb R$ it happens that $\det(\alpha_i(v_j)_{i,j=1,\ldots,n})\neq0$, then $v_1,\ldots,v_n$ must be $\Bbb R$-linearly independent. The reason is that any $\Bbb R$-linear dependence among the $v_j$ would imply the same linear dependence of the columns of this matrix, and therefore the vanishing of its determinant. However, vanishing of the determinant does not conversely imply linear dependence of the vectors $v_i$ (many other circumstances can cause vanishing of the determinant, for instance a linear dependency between the forms $\alpha_i$ would do so, or if some $v_j$ should be in the kernel of every $\alpha_i$.)

In the case where the vectors $v_j$ are actually functions$~f_j$ on$~\Bbb R$, the linear forms could be evaluation of the functions in points $a_1,\ldots,a_n\in\Bbb R$, and the determinant would look like $$ \left| \begin{matrix} f_1(a_1)&f_2(a_1)&\ldots&f_n(a_1)\\ f_1(a_2)&f_2(a_2)&\ldots&f_n(a_2)\\ \vdots & \vdots & \ddots & \vdots \\ f_1(a_n)&f_2(a_n)&\ldots&f_n(a_n)\\ \end{matrix} \right|. $$ Note that this is a matrix with real entries, and the determinant is computed in$~\Bbb R$. Or, supposing the functions are sufficiently differentiable, one could instead choose a single point$~a\in\Bbb R$ at which one evaluates derivatives of the functions: $$ \left| \begin{matrix} f_1(a)&f_2(a)&\ldots&f_n(a)\\ f_1'(a)&f_2'(a)&\ldots&f_n'(a)\\ \vdots & \vdots & \ddots & \vdots \\ f_1^{(n-1)}(a)&f_2^{(n-1)}(a)&\ldots&f_n^{(n-1)}(a)\\ \end{matrix} \right|. $$ Finally, to increase the chances of of finding a nonvanishing determinant, one may do the latter while varying the point$~a$, to obtain a function $$ t\mapsto\left| \begin{matrix} f_1(t)&f_2(t)&\ldots&f_n(t)\\ f_1'(t)&f_2'(t)&\ldots&f_n'(t)\\ \vdots & \vdots & \ddots & \vdots \\ f_1^{(n-1)}(t)&f_2^{(n-1)}(t)&\ldots&f_n^{(n-1)}(t)\\ \end{matrix} \right|. $$ This is not the determinant of a matrix with functions as entries, but a function with real values (which are computed as determinant of a real matrix), indeed this function is the Wronskian of $f_1,\ldots,f_n$. If the function has any nonzero value this will prove $f_1,\ldots,f_n$ are linearly independent; however we still cannot conversely conclude that the determinant vanishing everywhere must be due to $\Bbb R$-linear dependence of the functions. Note that this is not in contradiction with the proposition above, since we are not talking about a matrix with entries in some ring$~R$ of functions. Note that the matrix $A$ of the question is not a Wronskian, and that it is singular for a quite trivial reason (having a zero row), and that nothing can be concluded about $\Bbb R$-linear dependence of the functions $t\mapsto t$ and $t\mapsto t^2$ (but these functions are not the columns of$~A$, nor are they even part of those columns).

(At the risk of causing more confusion, I'll conclude by saying that if we had been talking about a matrix with functions as entries, then the proposition would ensure that when the determinant gives the zero function, there must be some $R$-linear dependence of the columns, in other words a linear dependence relation where functions are (also) allowed as coefficients of the relation.)


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